M LHSD CA 12e ch02a · 2018-09-16 · The ,; , . −. The ,; ,

Copyright © 2017 Pearson Education, Inc. 170

Chapter 2 GRAPHS AND FUNCTIONS

Section 2.1 Rectangular Coordinates and Graphs

1. The point (–1, 3) lies in quadrant II in the rectangular coordinate system.

2. The point (4, 6) lies on the graph of the equation y = 3x – 6. Find the y-value by letting x = 4 and solving for y.

3 4 6 12 6 6y

3. Any point that lies on the x-axis has y-coordinate equal to 0.

4. The y-intercept of the graph of y = –2x + 6 is (0, 6).

5. The x-intercept of the graph of 2x + 5y = 10 is (5, 0). Find the x-intercept by letting y = 0 and solving for x.

2 5 0 10 2 10 5x x x

6. The distance from the origin to the point (–3, 4) is 5. Using the distance formula, we have

2 2

2 2

( , ) ( 3 0) (4 0)

3 4 9 16 25 5

d P Q

7. True

8. True

9. False. The midpoint of the segment joining (0, 0) and (4, 4) is

4 0 4 0 4 4, , 2, 2 .

2 2 2 2

10. False. The distance between the point (0, 0) and (4, 4) is

2 2 2 2( , ) (4 0) (4 0) 4 4

16 16 32 4 2

d P Q

11. Any three of the following: 2, 5 , 1,7 , 3, 9 , 5, 17 , 6, 21

12. Any three of the following: 3,3 , 5, 21 , 8,18 , 4,6 , 0, 6

13. Any three of the following: (1999, 35), (2001, 29), (2003, 22), (2005, 23), (2007, 20), (2009, 20)

14. Any three of the following:

2002,86.8 , 2004, 89.8 , 2006, 90.7 ,

2008, 97.4 , 2010, 106.5 , 2012,111.4 ,

2014, 111.5

15. P(–5, –6), Q(7, –1)

(a) 2 2

2 2

( , ) [7 – (–5)] [ 1 – (–6)]

12 5 169 13

d P Q

(b) The midpoint M of the segment joining points P and Q has coordinates

–5 7 6 ( 1) 2 7, ,

2 2 2 27

1, .2

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Section 2.1 Rectangular Coordinates and Graphs 171

Copyright © 2017 Pearson Education, Inc.

16. P(–4, 3), Q(2, –5)

(a) 2 2

2 2

( , ) [2 – (– 4)] (–5 – 3)

6 (–8) 100 10

d P Q


– 4 2 3 (–5) 2 2, ,

2 2 2 21, 1 .

17. (8, 2), (3, 5)P Q

(a)

2 2

2 2

( , ) (3 – 8) (5 – 2)

5 3

25 9 34

d P Q


8 3 2 5 11 7, ,

2 2 2 2

.

18. P (−8, 4), Q (3, −5)

(a) 2 2

2 2

( , ) 3 – 8 5 4

11 (–9) 121 81

202

d P Q


–8 3 4 (–5) 5 1, , .

2 2 2 2

19. P(–6, –5), Q(6, 10)

(a) 2 2

2 2

( , ) [6 – (– 6)] [10 – (–5)]

12 15 144 225

369 3 41

d P Q


– 6 6 –5 10 0 5 5, , 0, .

2 2 2 2 2

20. P(6, –2), Q(4, 6)

(a) 2 2

2 2

( , ) (4 – 6) [6 – (–2)]

(–2) 8

4 64 68 2 17

d P Q


6 4 2 6 10 4, , 5, 2

2 2 2 2

21. 3 2, 4 5 ,P 2, – 5Q

(a)

2 2

2 2

( , )

2 – 3 2 – 5 – 4 5

–2 2 –5 5

8 125 133

d P Q


3 2 2 4 5 (– 5),

2 2

4 2 3 5 3 5, 2 2, .

2 2 2

22. – 7, 8 3 , 5 7, – 3P Q

(a) 2 2

2 2

( , )

[5 7 – (– 7)] (– 3 – 8 3)

(6 7) (–9 3) 252 243

495 3 55

d P Q


– 7 5 7 8 3 (– 3),

2 2

4 7 7 3 7 3, 2 7, .

2 2 2

23. Label the points A(–6, –4), B(0, –2), and C(–10, 8). Use the distance formula to find the length of each side of the triangle.

2 2

2 2

( , ) [0 – (– 6)] [–2 – (– 4)]

6 2 36 4 40

d A B

2 2

2 2

( , ) (–10 – 0) [8 – (–2)]

( 10) 10 100 100

200

d B C

2 2

2 2

( , ) [–10 – (– 6)] [8 – (– 4)]

(– 4) 12 16 144 160

d A C

Because 2 2 240 160 200 ,

triangle ABC is a right triangle.

172 Chapter 2 Graphs and Functions


24. Label the points A(–2, –8), B(0, –4), and C(–4, –7). Use the distance formula to find the length of each side of the triangle.

2 2

2 2

( , ) [0 – (–2)] [– 4 – (–8)]

2 4 4 16 20

d A B

2 2

2 2

( , ) (– 4 – 0) [–7 – (– 4)]

(– 4) (–3) 16 9

25 5

d B C

2 2

2 2

( , ) [– 4 – (–2)] [–7 – (–8)]

(–2) 1 4 1 5

d A C

Because 2 2 2( 5) ( 20) 5 20 25 5 , triangle ABC is a right triangle.

25. Label the points A(–4, 1), B(1, 4), and C(–6, –1).

2 2

2 2

( , ) [1 – (– 4)] (4 – 1)

5 3 25 9 34

d A B

2 2

2 2

( , ) (– 6 –1) (–1 – 4)

(–7) (–5) 49 25 74

d B C

2 2

2 2

( , ) [– 6 – (– 4)] (–1 –1)

(–2) (–2) 4 4 8

d A C

Because 2 2 2( 8) ( 34) ( 74) because

8 34 42 74, triangle ABC is not a right triangle.

26. Label the points A(–2, –5), B(1, 7), and C(3, 15).

2 2

2 2

( , ) [1 ( 2)] [7 ( 5)]

3 12 9 144 153

d A B

2 2

2 2

( , ) (3 1) (15 7)

2 8 4 64 68

d B C

2 2

2 2

( , ) [3 ( 2)] [15 ( 5)]

5 20 25 400 425

d A C

Because 22 2( 68) ( 153) 425 because

68 153 221 425 , triangle ABC is not a right triangle.


2 2

2 2

( , ) 2 – –4 5 3

6 2 36 4 40

d A B

2 2

2 2

( , ) 1 2 6 5

(– 3) (–11)

9 121 130

d B C

2 2

22

( , ) –1 – –4 6 3

3 9 9 81 90

d A C

Because 2 2 240 90 130 , triangle

ABC is a right triangle.

28. Label the points A(–7, 4), B(6, –2), and C(0, –15).

2 2

22

( , ) 6 – –7 2 4

13 6

169 36 205

d A B

22

2 2

( , ) 0 6 –15 – –2

– 6 –13

36 169 205

d B C

2 2

22

( , ) 0 – –7 15 4

7 19 49 361 410

d A C

Because 2 2 2205 205 410 ,


29. Label the given points A(0, –7), B(–3, 5), and C(2, –15). Find the distance between each pair of points.

22

2 2

( , ) 3 0 5 – –7

–3 12 9 144

153 3 17

d A B

2 2

22

( , ) 2 – 3 –15 – 5

5 –20 25 400

425 5 17

d B C

22

22

( , ) 2 0 15 – 7

2 –8 68 2 17

d A C

Because ( , ) ( , ) ( , )d A B d A C d B C or

3 17 2 17 5 17, the points are collinear.



30. Label the points A(–1, 4), B(–2, –1), and C(1, 14). Apply the distance formula to each pair of points.

2 2

2 2

( , ) –2 – –1 –1 – 4

–1 –5 26

d A B

2 2

2 2

( , ) 1 – –2 14 – –1

3 15 234 3 26

d B C

2 2

2 2

( , ) 1 – –1 14 – 4

2 10 104 2 26

d A C

Because 26 2 26 3 26 , the points are collinear.

31. Label the points A(0, 9), B(–3, –7), and C(2, 19).

2 2

2 2

( , ) (–3 – 0) (–7 – 9)

(–3) (–16) 9 256

265 16.279

d A B

2 2

2 2

( , ) 2 – –3 19 – –7

5 26 25 676

701 26.476

d B C

2 2

2 2

( , ) 2 – 0 19 – 9

2 10 4 100

104 10.198

d A C

Because ( , ) ( , ) ( , )d A B d A C d B C

or 265 104 70116.279 10.198 26.476,

26.477 26.476,

the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.)

32. Label the points A(–1, –3), B(–5, 12), and C(1, –11).

2 2

2 2

( , ) –5 – –1 12 – –3

– 4 15 16 225

241 15.5242

d A B

2 2

22

( , ) 1 – –5 –11 –12

6 –23 36 529

565 23.7697

d B C

2 2

22

( , ) 1 – –1 –11 – –3

2 –8 4 64

68 8.2462

d A C

Because d(A, B) + d(A, C) d(B, C)

or 241 68 56515.5242 8.2462 23.7697

23.7704 23.7697,


33. Label the points A(–7, 4), B(6,–2), and C(–1,1).

2 2

22

( , ) 6 – –7 2 4

13 6 169 36

205 14.3178

d A B

22

2 2

( , ) 1 6 1 –2

7 3 49 9

58 7.6158

d B C

2 2

22

( , ) 1 – –7 1 4

6 –3 36 9

45 6.7082

d A C

Because d(B, C) + d(A, C) d(A, B) or

58 45 2057.6158 6.7082 14.3178

14.3240 14.3178,


34. Label the given points A(–4, 3), B(2, 5), and C(–1, 4). Find the distance between each pair of points.

2 2 2 2( , ) 2 – –4 5 3 6 2

36 4 40 2 10

d A B

2 2

2 2

( , ) ( 1 2) (4 5)

3 (–1) 9 1 10

d B C

2 2

2 2

( , ) 1 – 4 4 3

3 1 9 1 10

d A C

Because ( , ) ( , ) ( , )d B C d A C d A B or

10 10 2 10, the points are collinear.



35. Midpoint (5, 8), endpoint (13, 10) 13 10

5 and 82 2

13 10 and 10 16

–3 and 6.

x y

x yx y

The other endpoint has coordinates (–3, 6).

36. Midpoint (–7, 6), endpoint (–9, 9) –9 9

–7 and 62 2

–9 –14 and 9 12

–5 and 3.

x y

x yx y



12 and 62 2

19 24 and 16 12

5 and – 4.

x y

x yx y

The other endpoint has coordinates (5, –4).


–9 and 82 2

–16 –18 and 9 16

–2 and 7

x y

x yx y


39. Midpoint (a, b), endpoint (p, q)

and2 2

2 and 2

2 and 2

p x q ya b

p x a q y bx a p y b q

The other endpoint has coordinates (2 , 2 )a p b q .

40. Midpoint 6 , 6a b , endpoint 3 , 5a b

3 56 and 6

2 23 12 and 5 12

9 and 7

a x b ya b

a x a b y bx a y b

The other endpoint has coordinates (9a, 7b).

41. The endpoints of the segment are (1990, 21.3) and (2012, 30.1).

1990 2012 21.3 30.9,

2 22001, 26.1

M

The estimate is 26.1%. This is very close to the actual figure of 26.2%.

42. The endpoints are (2006, 7505) and (2012, 3335)

2006 2012 7505 3335,

2 22009, 5420

M

According to the model, the average national advertising revenue in 2009 was $5420 million. This is higher than the actual value of $4424 million.

43. The points to use are (2011, 23021) and (2013, 23834). Their midpoint is

2011 2013 23,021 23,834,

2 2(2012, 23427.5).

In 2012, the poverty level cutoff was approximately $23,428.

44. (a) To estimate the enrollment for 2003, use the points (2000, 11,753) and (2006, 13,180)

2000 2006 11,753 13,180,

2 22003, 12466.5

M

The enrollment for 2003 was about 12,466.5 thousand.

(b) To estimate the enrollment for 2009, use the points (2006, 13,180) and (2012, 14,880)

2006 2012 13,180 14,880,

2 22009, 14030

M

The enrollment for 2009 was about 14,030 thousand.

45. The midpoint M has coordinates

1 2 1 2, .2 2

x x y y

2 21 2 1 2

1 1

2 21 2 1 1 2 1

2 22 1 2 1

2 22 1 2 1

2 22 1 2 1

2 212 1 2 12

( , )

– –2 2

2 2– –

2 2 2 2

2 2

4 4

4

d P M

x x y yx y

x x x y y y

x x y y

x x y y

x x y y

x x y y

(continued on next page)



(continued)

2 21 2 1 2

2 2

2 22 1 2 2 1 2

2 22 1 2 1

2 22 1 2 1

2 22 1 2 1

2 212 1 2 12

( , )

2 2

2 2

2 2 2 2

2 2

4 4

4

d M Q

x x y yx y

x x x y y y

x x y y

x x y y

x x y y

x x y y

2 22 1 2 1( , )d P Q x x y y

Because

2 212 1 2 12

2 212 1 2 12

2 22 1 2 1 ,

x x y y

x x y y

x x y y

this shows ( , ) ( , ) ( , )d P M d M Q d P Q and

( , ) ( , ).d P M d M Q

46. The distance formula, 2 2

2 1 2 1( – ) ( – ) ,d x x y y can be written

as 2 2 1/ 22 1 2 1[( – ) ( – ) ] .d x x y y

In exercises 47−58, other ordered pairs are possible.

47. (a) x y 0 −2 y-intercept:

12

00 2 2

xy

4 0 x-intercept:

1212

00 2

2 4

yxx x

2 −1 additional point

(b)

48. (a) x y 0 2 y-intercept:

0

10 2 2

2

x

y

4 0 x-intercept: 0

10 2

21

2 42

y

x

x x

2 1 additional point

(b)

49. (a) x y 0 5

3 y-intercept:

53

02 0 3 5

3 5

xy

y y

52

0 x-intercept:

52

02 3 0 5

2 5

yxx x


(b)




0

3 0 2 62 6 3

xy

y y

2 0 x-intercept:

0

3 2 0 63 6 2

yxx x


(b)

51. (a) x y 0 0 x- and y-intercept:

20 0


−2 4 additional point

(b)


2

0

0 20 2 2

xyy y

−1 3 additional point 2 6 additional point

no x-intercept: 2

20 0 2

2 2

y xx x

(b)

53. (a) x y 3 0 x-intercept:

0

0 30 3 3

yx

x x

4 1 additional point 7 2 additional point

no y-intercept:

0 0 3 3x y y

(b)


0

0 30 3 3

xyy y


9 0 x-intercept: 0

0 3

3 9

yxx x

(b)




00 2

2 2

xyy y

2 0 x-intercept: 0

0 20 2 2

yxx x



(b)

56. (a) x y −2 −2 additional point −4 0 x-intercept:

00 4

0 40 4 4

yx

xx x

0 −4 y-intercept: 0

0 4

4 4

xyy y

(b)


30 0 −1 −1 additional point 2 8 additional point

(b)


30 0 1 −1 additional point2 −8 additional point

(b)

59. Points on the x-axis have y-coordinates equal to 0. The point on the x-axis will have the same x-coordinate as point (4, 3). Therefore, the line will intersect the x-axis at (4, 0).

60. Points on the y-axis have x-coordinates equal to 0. The point on the y-axis will have the same y-coordinate as point (4, 3). Therefore, the line will intersect the y-axis at (0, 3).

61. Because (a, b) is in the second quadrant, a is negative and b is positive. Therefore, (a, – b) will have a negative x–coordinate and a negative y-coordinate and will lie in quadrant III. (–a, b) will have a positive x-coordinate and a positive y-coordinateand will lie in quadrant I. (–a, – b) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV. (b, a) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV.



62. Label the points ( 2, 2),A (13,10),B

(21, 5),C and (6, 13).D To determine which points form sides of the quadrilateral (as opposed to diagonals), plot the points.

Use the distance formula to find the length of

each side.

2 2

2 2

2 2

22

( , ) 13 2 10 2

15 8 225 64

289 17

( , ) 21 13 5 10

8 15 64 225

289 17

d A B

d B C

22

2 2

( , ) 6 21 13 5

15 8

225 64 289 17

d C D

22

2 2

( , ) 2 6 2 13

8 15

64 225 289 17

d D A

Because all sides have equal length, the four points form a rhombus.

63. To determine which points form sides of the quadrilateral (as opposed to diagonals), plot the points.


each side.

2 2

2 2

( , ) 5 1 2 1

4 1 16 1 17

d A B

2 2

2 2

( , ) 3 5 4 2

2 2 4 4 8

d B C

2 2

2 2

( , ) 1 3 3 4

4 1

16 1 17

d C D

2 2

22

( , ) 1 1 1 3

2 2 4 4 8

d D A

Because d(A, B) = d(C, D) and d(B, C) = d(D, A), the points are the vertices of a parallelogram. Because d(A, B) ≠ d(B, C), the points are not the vertices of a rhombus.

64. For the points A(4, 5) and D(10, 14), the difference of the x-coordinates is 10 – 4 = 6 and the difference of the y-coordinates is 14 – 5 = 9. Dividing these differences by 3, we obtain 2 and 3, respectively. Adding 2 and 3 to the x and y coordinates of point A, respectively, we obtain B(4 + 2, 5 + 3) or B(6, 8).

Adding 2 and 3 to the x- and y- coordinates of point B, respectively, we obtain C(6 + 2, 8 + 3) or C(8, 11). The desired points are B(6, 8) and C(8, 11).

We check these by showing that d(A, B) = d(B, C) = d(C, D) and that d(A, D) = d(A, B) + d(B, C) + d(C, D).

2 2

2 2

2 2

2 2

( , ) 6 4 8 5

2 3 4 9 13

( , ) 8 6 11 8

2 3 4 9 13

d A B

d B C

2 2

2 2

2 2

2 2

( , ) 10 8 14 11

2 3 4 9 13

( , ) 10 4 14 5

6 9 36 81

117 9(13) 3 13

d C D

d A D

d(A, B), d(B, C), and d(C, D) all have the same measure and d(A, D) = d(A, B) + d(B, C) + d(C, D) Because

3 13 13 13 13.

Section 2.2 Circles 179


Section 2.2 Circles

1. The circle with equation 2 2 49x y has center with coordinates (0, 0) and radius equal to 7.

2. The circle with center (3, 6) and radius 4 has

equation 23 6 16.x y

3. The graph of 2 24 7 9x y has center

with coordinates (4, –7).

4. The graph of 22 5 9x y has center with

coordinates (0, 5).

5. This circle has center (3, 2) and radius 5. This is graph B.

6. This circle has center (3, –2) and radius 5. This is graph C.

7. This circle has center (–3, 2) and radius 5. This is graph D.

8. This circle has center (–3, –2) and radius 5. This is graph A.

9. The graph of 2 2 0x y has center (0, 0) and radius 0. This is the point (0, 0). Therefore, there is one point on the graph.

10. 100 is not a real number, so there are no

points on the graph of 2 2 100.x y

11. (a) Center (0, 0), radius 6

2 2

2 2 2 2 2

0 0 6

0 0 6 36

x y

x y x y

(b)


2 2

2 2 2 2 2

0 0 9

0 0 9 81

x y

x y x y

(b)


2 2

2 2 2

2 2

2 0 6

2 0 6

( – 2) 36

x y

x yx y

(b)


2 2

2 2

3 0 3

3 9

x yx y

(b)


2 2

22

0 4 4

4 16

x yx y

(b)



16. (a) Center (0, –3), radius 7

22

22 2

2 2

0 3 7

0 3 7

( 3) 49

x y

x yx y

(b)

17. (a) Center (–2, 5), radius 4

2 2

2 2 2

2 2

2 5 4

[ – (–2)] ( – 5) 4

( 2) ( – 5) 16

x y

x yx y

(b)


2 2

2 2 2

2 2

4 3 5

4 3 5

4 3 25

x y

x yx y

(b)


22

2 2 2

2 2

5 4 7

( – 5) [ – (– 4)] 7

( – 5) ( 4) 49

x y

x yx y

(b)

20. (a) Center (–3, –2), radius 6

2 2

2 2 2

2 2

3 2 6

3 2 6

( 3) ( 2) 36

x y

x yx y

(b)

21. (a) Center 2, 2 , radius 2

2 2

2 2

2 2 2

2 2 2

x y

x y

(b)


2 2

2 2 2

2 2

3 3 3

3 3 3

3 3 3

x y

x y

x y



(b)

23. (a) The center of the circle is located at the midpoint of the diameter determined by the points (1, 1) and (5, 1). Using the midpoint formula, we have

1 5 1 1, 3,1

2 2C

. The radius is

one-half the length of the diameter:

2 215 1 1 1 2

2r

The equation of the circle is

2 23 1 4x y

(b) Expand 2 23 1 4x y to find the

equation of the circle in general form:

2 2

2 2

2 2

3 1 46 9 2 1 4

6 2 6 0

x yx x y y

x y x y

24. (a) The center of the circle is located at the midpoint of the diameter determined by the points (−1, 1) and (−1, −5).

Using the midpoint formula, we have

1 ( 1) 1 ( 5), 1, 2 .

2 2C

The radius is one-half the length of the diameter:

2 211 1 5 1 3

2r


2 21 2 9x y



2 2

2 2

2 2

1 2 92 1 4 4 9

2 4 4 0

x yx x y y

x y x y

25. (a) The center of the circle is located at the midpoint of the diameter determined by the points (−2, 4) and (−2, 0). Using the midpoint formula, we have

2 ( 2) 4 0, 2, 2 .

2 2C


2 212 2 4 0 2

2r


2 22 2 4x y



2 2

2 2

2 2

2 2 44 4 4 4 4

4 4 4 0

x yx x y y

x y x y

26. (a) The center of the circle is located at the midpoint of the diameter determined by the points (0, −3) and (6, −3). Using the midpoint formula, we have

0 6 3 ( 3), 3, 3

2 2C

.


2216 0 3 3 3

2r


2 23 3 9x y



2 2

2 2

2 2

3 3 96 9 6 9 9

6 6 9 0

x yx x y y

x y x y

27. 2 2 6 8 9 0x y x y Complete the square on x and y separately.

2 2

2 2

2 2

6 8 –9

6 9 8 16 –9 9 16

3 4 16

x x y y

x x y y

x y

Yes, it is a circle. The circle has its center at (–3, –4) and radius 4.



28. 2 2 8 – 6 16 0x y x y Complete the square on x and y separately.

2 2

2 2

2 2

8 – 6 16

8 16 – 6 9 –16 16 9

4 – 3 9

x x y y

x x y y

x y

Yes, it is a circle. The circle has its center at (–4, 3) and radius 3.

29. 2 2 4 12 4x y x y Complete the square on x and y separately.

2 2

2 2

2 2

– 4 12 – 4

– 4 4 12 36 – 4 4 36

– 2 6 36

x x y y

x x y y

x y

Yes, it is a circle. The circle has its center at (2, –6) and radius 6.

30. 2 2 –12 10 –25x y x y Complete the square on x and y separately.

2 2

2 2

2 2

–12 10 –25

–12 36 10 25

– 25 36 25

– 6 5 36

x x y y

x x y y

x y


31. 2 24 4 4 –16 – 19 0x y x y Complete the square on x and y separately.

2 2

2 214

14

4 4 – 4 19

4 4 – 4 4

19 4 4 4

x x y y

x x y y

2 212

2 212

4 4 – 2 36

– 2 9

x y

x y

Yes, it is a circle with center 12

, 2 and

radius 3.

32. 2 29 9 12 – 18 – 23 0x y x y Complete the square on x and y separately.

2 243

2 24 43 9

49

9 9 – 2 23

9 9 – 2 1

23 9 9 1

x x y y

x x y y

2 223

2 223

9 9 –1 36

–1 4

x y

x y


, 1 and

radius 2.


2 2

2 2

2 2

2 – 6 –14

2 1 – 6 9 –14 1 9

1 – 3 – 4

x x y y

x x y y

x y

The graph is nonexistent.


2 2

2 2

2 2

4 – 8 –32

4 4 – 8 16

– 32 4 16

2 – 4 –12

x x y y

x x y y

x y



2 2

2 2

2 2

6 6 18

6 9 6 9 18 9 9

3 3 0

x x y y

x x y y

x y

The graph is the point (3, 3).


2 2

2 2

2 2

4 4 8

4 4 4 4 8 4 4

2 2 0

x x y y

x x y y

x y

The graph is the point (−2, −2).

37. 2 29 9 6 6 23 0x y x y Complete the square on x and y separately.

2 2

2 22 23 3

22 2 232 1 2 1 1 13 9 3 9 9 9 9

2 2 225 51 13 3 9 3

9 6 9 6 23

9 9 23

x x y y

x x y y

x x y y

x y

Yes, it is a circle with center 1 13 3

, and

radius 53

.




2 2

2 21 14 4

1 14 4

2 21 12 2

2 2 91 12 2 4

4 4 7

4 4 –

7 4 4

4 4 – 9

–

x x y y

x x y y

x y

x y


, and

radius 32

.

39. The equations of the three circles are 2 2( 7) ( 4) 25x y , 2 2( 9) ( 4) 169x y , and 2 2( 3) ( 9) 100x y . From the graph of the

three circles, it appears that the epicenter is located at (3, 1).

Check algebraically:

2 2

2 2

2 2

( 7) ( 4) 25

(3 7) (1 4) 25

4 3 25 25 25

x y

2 2

2 2

2 2

( 9) ( 4) 169

(3 9) (1 4) 169

12 5 169 169 169

x y

2 2

2 2

2 2

( 3) ( 9) 100

(3 3) (1 9) 100

6 ( 8) 100 100 100

x y

(3, 1) satisfies all three equations, so the epicenter is at (3, 1).

40. The three equations are 2 2( 3) ( 1) 5x y , 2 2( 5) ( 4) 36x y , and 2 2( 1) ( 4) 40x y . From the graph of the



2 2

2 2

2 2

( 3) ( 1) 5

(5 3) (2 1) 5

2 1 5 5 5

x y

2 2

2 2

2

( 5) ( 4) 36

(5 5) (2 4) 36

6 36 36 36

x y

2 2

2 2

2 2

( 1) ( 4) 40

(5 1) (2 4) 40

6 ( 2) 40 40 40

x y


41. From the graph of the three circles, it appears that the epicenter is located at (−2, −2).


2 2

2 2

2 2

( 2) ( 1) 25

( 2 2) ( 2 1) 25

( 4) ( 3) 2525 25

x y

2 2

2 2

2 2

( 2) ( 2) 16

( 2 2) ( 2 2) 16

0 ( 4) 1616 16

x y

2 2

2 2

2 2

( 1) ( 2) 9

( 2 1) ( 2 2) 9

( 3) 0 99 9

x y

(−2, −2) satisfies all three equations, so the epicenter is at (−2, −2).



42. From the graph of the three circles, it appears that the epicenter is located at (5, 0).


2 2

2 2

2 2

( 2) ( 4) 25

(5 2) (0 4) 25

3 ( 4) 2525 25

x y

2 2

2 2

2 2

( 1) ( 3) 25

(5 1) (0 3) 25

4 3 2525 25

x y

2 2

2 2

2 2

( 3) ( 6) 100

(5 3) (0 6) 100

8 6 100100 100

x y


43. The radius of this circle is the distance from the center C(3, 2) to the x-axis. This distance is 2, so r = 2.

2 2 2

2 2

( – 3) ( – 2) 2

( – 3) ( – 2) 4

x yx y

44. The radius is the distance from the center C(–4, 3) to the point P(5, 8).

2 2

2 2

[5 – (– 4)] (8 – 3)

9 5 106

r

The equation of the circle is 2 2 2

2 3

[ – (– 4)] ( – 3) ( 106)

( 4) ( – 3) 106

x yx y

45. Label the points P(x, y) and Q(1, 3).

If ( , ) 4d P Q , 2 21 3 4x y

2 21 3 16.x y

If x = y, then we can either substitute x for y or y for x. Substituting x for y we solve the following:

2 2

2 2

2

2

2

1 3 16

1 2 9 6 16

2 8 10 16

2 8 6 0

4 3 0

x xx x x x

x xx xx x

To solve this equation, we can use the quadratic formula with a = 1, b = –4, and c = –3.

24 4 4 1 3

2 1

4 16 12 4 28

2 2

4 2 72 7

2

x

Because x = y, the points are

2 7, 2 7 and 2 – 7, 2 7 .

46. Let P(–2, 3) be a point which is 8 units from Q(x, y). We have

2 2( , ) 2 3 8d P Q x y

2 22 3 64.x y

Because x + y = 0, x = –y. We can either substitute x for y or y for x. Substituting

x for y we solve the following:

22

2 2

2 2

2

2

2 3 64

2 3 64

4 4 9 6 64

2 10 13 64

2 10 51 0

x x

x xx x x x

x xx x

To solve this equation, use the quadratic formula with a = 2, b = 10, and c = –51.

210 10 4 2 51

2 2

10 100 408

410 4 12710 508

4 410 2 127 5 127

4 2

x

Because y x the points are

5 127 5 127,

2 2

and

–5 127 5 127, .

2 2



47. Let P(x, y) be a point whose distance from

A(1, 0) is 10 and whose distance from

B(5, 4) is 10 . d(P, A) = 10 , so

2 2(1 ) (0 ) 10x y 2 2(1 ) 10.x y ( , ) 10, sod P B

2 2(5 ) (4 ) 10x y 2 2(5 ) (4 ) 10.x y Thus,

2 2 2 2

2 2

2 2

(1 ) (5 ) (4 )

1 2

25 10 16 81 2 41 10 8

8 40 85

x y x yx x y

x x y yx x yy xy x

Substitute 5 – x for y in the equation 2 2(1 ) 10x y and solve for x. 2 2

2 2

2 2

2

(1 ) (5 ) 10

1 2 25 10 10

2 12 26 10 2 12 16 0

6 8 0 ( 2)( 4) 02 0 or 4 0

2 or 4

x xx x x x

x x x xx x x xx x

x x

To find the corresponding values of y use the equation y = 5 – x. If x = 2, then y = 5 – 2 = 3. If x = 4, then y = 5 – 4 = 1. The points satisfying the conditions are (2, 3) and (4, 1).

48. The circle of smallest radius that contains the points A(1, 4) and B(–3, 2) within or on its boundary will be the circle having points A and B as endpoints of a diameter. The center will be M, the midpoint:

1 3 4 2 2 6, ,

2 2 2 2

(−1, 3).

The radius will be the distance from M to either A or B:

2 2

2 2

( , ) [1 ( 1)] (4 3)

2 1 4 1 5

d M A


22 2

2 2

1 3 5

1 3 5.

x y

x y

49. Label the points A(3, y) and B(–2, 9). If d(A, B) = 12, then

2 2

2 2

2 2 2

2

2

2 3 9 12

5 9 12

5 9 12

25 81 18 144

18 38 0

y

y

yy y

y y

Solve this equation by using the quadratic formula with a = 1, b = –18, and c = –38:

218 18 4 1 38

2 1

18 324 152 18 476

2 1 2

y

18 4 119 18 2 119

9 1192 2

The values of y are 9 119 and 9 119 .

50. Because the center is in the third quadrant, the

radius is 2 , and the circle is tangent to both

axes, the center must be at ( 2, 2). Using the center-radius of the equation of a circle, we have

2 2 2

2 2

2 2 2

2 2 2.

x y

x y

51. Let P(x, y) be the point on the circle whose distance from the origin is the shortest. Complete the square on x and y separately to write the equation in center-radius form:

2 2

2 2

2 2

16 14 88 0

16 64 14 4988 64 49

( 8) ( 7) 25

x x y yx x y y

x y

So, the center is (8, 7) and the radius is 5.

2 2( , ) 8 7 113d C O . Because the

length of the radius is 5, ( , ) 113 5d P O .



52. Using compasses, draw circles centered at Wickenburg, Kingman, Phoenix, and Las Vegas with scaled radii of 50, 75, 105, and 180 miles respectively. The four circles should intersect at the location of Nothing.


3 –9–1+5 4 6 , , (2, – 3).

2 2 2 2

54. Use points C(2, –3) and P(–1, 3).

22

2 2

( , ) –1 – 2 3 – –3

–3 6 9 36

45 3 5

d C P

The radius is 3 5.

55. Use points C(2, –3) and Q(5, –9).

22

22

( , ) 5 – 2 –9 – –3

3 –6 9 36

45 3 5

d C Q

The radius is 3 5.

56. Use the points P(–1, 3) and Q(5, –9).

Because 2 2( , ) 5 – –1 –9 – 3d P Q

226 –12 36 144 180

6 5 ,the radius is 1

( , ).2

d P Q Thus

16 5 3 5.

2r

57. The center-radius form for this circle is 2 2 2( – 2) ( 3) (3 5)x y 2 2( – 2) ( 3) 45.x y

58. Label the endpoints of the diameter P(3, –5) and Q(–7, 3). The midpoint M of the segment joining P and Q has coordinates

3 (–7) –5 3 4 –2, , (–2, –1).

2 2 2 2

The center is C(–2, –1). To find the radius, we can use points C(–2, –1) and P(3, –5)

2 2

22

( , ) 3 – –2 –5 – –1

5 –4 25 16 41

d C P

We could also use points C(–2, –1).and Q(–7, 3).

2 2

2 2

( , ) 7 – –2 3 – –1

5 4 25 16 41

d C Q

We could also use points P(3, –5) and Q(–7, 3) to find the length of the diameter. The length of the radius is one-half the length of the diameter.

22

2 2

( , ) 7 – 3 3 – –5

10 8 100 64

164 2 41

d P Q

1 1( , ) 2 41 41

2 2d P Q

The center-radius form of the equation of the circle is

2 2 2

2 2

[ – (–2)] [ – (–1)] ( 41)

( 2) ( 1) 41

x yx y

59. Label the endpoints of the diameter P(–1, 2) and Q(11, 7). The midpoint M of the segment joining P and Q has coordinates

1 11 2 7 9, 5, .

2 2 2

The center is 92

5, .C To find the radius, we

can use points 92

5,C and P(−1, 2).

22 92

22 5 169 132 4 2

( , ) 5 – –1 2

6

d C P

We could also use points 92

5,C and

Q(11, 7).

22 92

22 5 169 132 4 2

( , ) 5 11 – 7

6

d C Q


Section 2.3 Functions 187


(continued)

Using the points P and Q to find the length of the diameter, we have

2 2

2 2

, 1 11 2 7

12 5

169 13

d P Q

1 1 13, 13

2 2 2d P Q


2 22 9 132 2

22 9 1692 4

5

5

x y

x y

60. Label the endpoints of the diameter P(5, 4) and Q(−3, −2). The midpoint M of the segment joining P and Q has coordinates

5 ( 3) 4 ( 2), 1, 1 .

2 2

The center is C(1, 1). To find the radius, we can use points C(1, 1) and P(5, 4).

2 2

2 2

( , ) 5 1 4 1

4 3 25 5

d C P

We could also use points C(1, 1) and Q(−3, −2).

2 2

2 2

( , ) 1 3 1 –2

4 3 25 5

d C Q


2 2

2 2

, 5 3 4 2

8 6 100 10

d P Q

1 1( , ) 10 5

2 2d P Q


2 2 2

2 2

1 1 5

1 1 25

x yx y

61. Label the endpoints of the diameter P(1, 4) and Q(5, 1). The midpoint M of the segment joining P and Q has coordinates

52

1 5 4 1, 3, .

2 2

The center is 52

3, .C

The length of the diameter PQ is

2 2 2 21 5 4 1 4 3 25 5.

The length of the radius is 512 2

5 .


2 22 5 52 2

22 5 252 4

3

3

x y

x y

62. Label the endpoints of the diameter P(−3, 10) and Q(5, −5). The midpoint M of the segment joining P and Q has coordinates

52

3 5 10 ( 5), 1, .

2 2

The center is 52

1, .C


22 2 23 5 10 5 8 15

289 17.


17 .


2 22 5 172 2

22 5 2892 4

1

1

x y

x y

Section 2.3 Functions

1. The domain of the relation

3,5 , 4,9 , 10,13 is 3, 4,10 .

2. The range of the relation in Exercise 1 is

5,9,13 .

3. The equation y = 4x – 6 defines a function with independent variable x and dependent variable y.

4. The function in Exercise 3 includes the ordered pair (6, 18).

5. For the function 4 2,f x x

2 4 2 2 8 2 10.f

6. For the function ,g x x 9 9 3.g

7. The function in Exercise 6, ,g x x has

domain 0, .




range 0, .

For exercises 9 and 10, use this graph.

9. The largest open interval over which the function graphed here increases is , 3 .

10. The largest open interval over which the function graphed here decreases is 3, .

11. The relation is a function because for each different x-value there is exactly one y-value. This correspondence can be shown as follows.


13. Two ordered pairs, namely (2, 4) and (2, 6), have the same x-value paired with different y-values, so the relation is not a function.

14. Two ordered pairs, namely (9, −2) and (9, 1), have the same x-value paired with different y-values, so the relation is not a function.





19. Two sets of ordered pairs, namely (1, 1) and (1, −1) as well as (2, 4) and (2, −4), have the same x-value paired with different y-values, so the relation is not a function. domain: {0, 1, 2}; range: {−4, −1, 0, 1, 4}

20. The relation is not a function because the x-value 3 corresponds to two y-values, 7 and 9. This correspondence can be shown as follows.

domain: {2, 3, 5}; range: {5, 7, 9, 11}

21. The relation is a function because for each different x-value there is exactly one y-value. domain: {2, 3, 5, 11, 17}; range: {1, 7, 20}

22. The relation is a function because for each different x-value there is exactly one y-value. domain: {1, 2, 3, 5}; range: {10, 15, 19, 27}


Domain: {0, −1, −2}; range: {0, 1, 2}




Domain: {0, 1, 2}; range: {0, −1, −2}

25. The relation is a function because for each different year, there is exactly one number for visitors. domain: {2010, 2011, 2012, 2013} range: {64.9, 63.0, 65.1, 63.5}

26. The relation is a function because for each basketball season, there is only one number for attendance. domain: {2011, 2012, 2013, 2014} range: {11,159,999, 11,210,832, 11,339,285, 11,181,735}

27. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: , ; range: ,

28. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: , ; range: , 4

29. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y. domain: 3, ; range: ,

30. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y. domain: [−4, 4]; range: [−3, 3]


32. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: [−2, 2]; range: [0, 4]

33. 2y x represents a function because y is always found by squaring x. Thus, each value of x corresponds to just one value of y. x can be any real number. Because the square of any real number is not negative, the range would be zero or greater.

domain: , ; range: 0,

34. 3y x represents a function because y is always found by cubing x. Thus, each value of x corresponds to just one value of y. x can be any real number. Because the cube of any real number could be negative, positive, or zero, the range would be any real number.

domain: , ; range: ,

35. The ordered pairs (1, 1) and (1, −1) both

satisfy 6.x y This equation does not represent a function. Because x is equal to the sixth power of y, the values of x are nonnegative. Any real number can be raised to the sixth power, so the range of the relation is all real numbers.

domain: 0, range: ,




satisfy 4.x y This equation does not represent a function. Because x is equal to the fourth power of y, the values of x are nonnegative. Any real number can be raised to the fourth power, so the range of the relation is all real numbers.

domain: 0, range: ,

37. 2 5y x represents a function because y is found by multiplying x by 2 and subtracting 5. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers. Because y is twice x, less 5, y also may be any real number, and so the range is also all real numbers.


38. 6 4y x represents a function because y is found by multiplying x by −6 and adding 4. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers. Because y is −6 times x, plus 4, y also may be any real number, and so the range is also all real numbers.


39. By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into x + y < 3 corresponds to many values of y. The ordered pairs (1, –2), (1, 1), (1, 0), (1, −1), and so on, all satisfy the inequality. Note that the points on the graphed line do not satisfy the inequality and only indicate the boundary of the solution set. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers.


40. By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into x − y < 4 corresponds to many values of y. The ordered pairs (1, −1), (1, 0), (1, 1), (1, 2), and so on, all satisfy the inequality. Note that the points on the graphed line do not satisfy the inequality and only indicate the boundary of the solution set. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers.




41. For any choice of x in the domain of ,y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have 0.x Because the radical is nonnegative, the range is also zero or greater.

domain: 0, ; range: 0,

42. For any choice of x in the domain of

,y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have 0.x The outcome of the radical is nonnegative, when you change the sign (by multiplying by −1), the range becomes nonpositive. Thus the range is zero or less.

domain: 0, ; range: , 0

43. Because 2xy can be rewritten as 2 ,xy

we can see that y can be found by dividing x into 2. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 0. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.

domain: , 0 0, ;

range: , 0 0,

44. Because xy = −6 can be rewritten as 6 ,xy

we can see that y can be found by dividing x into −6. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 0. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.

domain: , 0 0, ;

range: , 0 0,


4 1y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have

4 1 0 4 1x x 14

.x Because the

radical is nonnegative, the range is also zero or greater.

domain: 1

4, ; range: 0,





7 2 0 2 7x x 7 72 2

or .x x

Because the radical is nonnegative, the range is also zero or greater.

domain: 72

, ; range: 0,

47. Given any value in the domain of 23

,xy we

find y by subtracting 3, then dividing into 2. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 3. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.

domain: ,3 3, ;

range: , 0 0,


,xy we

find y by subtracting 5, then dividing into −7. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 5. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.

domain: ,5 5, ;

range: , 0 0,

49. B. The notation f(3) means the value of the dependent variable when the independent variable is 3.

50. Answers will vary. An example is: The cost of gasoline depends on the number of gallons used; so cost is a function of number of gallons.

51.

3 4

0 3 0 4 0 4 4

f x xf

52.

3 4

3 3 3 4 9 4 13

f x xf

53.

2

2

4 1

2 2 4 2 1

4 8 1 11

g x x xg

54.

2

2

4 1

10 10 4 10 1100 40 1 59

g x x xg

55. 1 1

3 3

3 4

3 4 1 4 3

f x xf

56. 7 7

3 3

3 4

3 4 7 4 11

f x xf

57.

2

21 1 12 2 2

1 114 4

4 1

4 1

2 1

g x x x

g

58.

2

21 1 14 4 4

1 116 16

4 1

4 1

1 1

g x x x

g

59.

3 4

3 4

f x xf p p



60.

2

2

4 1

4 1

g x x xg k k k

61.

3 4

3 4 3 4

f x xf x x x

62.

2

2

2

4 1

4 1

4 1

g x x xg x x x

x x

63.

3 4

2 3 2 43 6 4 3 2

f x xf x x

x x

64.

3 4

4 3 4 43 12 4 3 8

f x xf a a

a a

65.

3 4

2 3 3 2 3 46 9 4 6 13

f x xf m m

m m

66.

3 4

3 2 3 3 2 49 6 4 9 10

f x xf t t

t t

67. (a) 2 2f (b) 1 3f

68. (a) 2 5f (b) 1 11f

69. (a) 2 15f (b) 1 10f

70. (a) 2 1f (b) 1 7f

71. (a) 2 3f (b) 1 3f

72. (a) 2 3f (b) 1 2f

73. (a) 2 0f (b) 0 4f

(c) 1 2f (d) 4 4f

74. (a) 2 5f (b) 0 0f

(c) 1 2f (d) 4 4f

75. (a) 2 3f (b) 0 2f

(c) 1 0f (d) 4 2f

76. (a) 2 3f (b) 0 3f

(c) 1 3f (d) 4 3f

77. (a)

1 13 3

3 123 12

12

34 4

x yy x

xy

y x f x x

(b) 13

3 3 4 1 4 3f

78. (a)

1 14 4

4 88 4

8 48

42 2

x yx y

x yx y

y x f x x

(b) 3 3 8 514 4 4 4 4

3 3 2 2f

79. (a)

2

2

2

2 3

2 3

2 3

y x xy x x

f x x x

(b) 23 2 3 3 32 9 3 3 18

f

80. (a)

2

2

2

3 2

3 2

3 2

y x xy x x

f x x x

(b) 23 3 3 3 23 9 3 2 32

f

81. (a)

8 84 43 3 3 3

4 3 84 3 8

4 8 34 8

3

x yx y

x yx y

y x f x x

(b) 8 84 12 43 3 3 3 3

3 3f

82. (a)

9 92 25 5 5 5

2 5 95 2 9

2 9

5

x yy x

xy

y x f x x

(b) 9 6 9 1525 5 5 5 5

3 3 3f

83. 3 4f



84. Because 20.2 0.2 3 0.2 1f

= 0.04 + 0.6 + 1 = 1.64, the height of the rectangle is 1.64 units. The base measures 0.3 − 0.2 = 0.1 unit. Because the area of a rectangle is base times height, the area of this rectangle is 0.1(1.64) = 0.164 square unit.

85. 3f is the y-component of the coordinate,

which is −4.


which is −3.

87. (a) 2, 0 (b) , 2

(c) 0,

88. (a) 3, 1 (b) 1,

(c) , 3

89. (a) , 2 ; 2,

(b) 2, 2 (c) none

90. (a) 3,3 (b) , 3 ; 3,

(c) none

91. (a) 1, 0 ; 1,

(b) , 1 ; 0, 1

(c) none

92. (a) , 2 ; 0, 2

(b) 2, 0 ; 2,

(c) none

93. (a) Yes, it is the graph of a function.

(b) [0, 24]

(c) When t = 8, y = 1200 from the graph. At 8 A.M., approximately 1200 megawatts is being used.

(d) The most electricity was used at 17 hr or 5 P.M. The least electricity was used at 4 A.M.

(e) 12 1900f

At 12 noon, electricity use is about 1900 megawatts.

(f) increasing from 4 A.M. to 5 P.M.; decreasing from midnight to 4 A.M. and from 5 P.M. to midnight

94. (a) At t = 2, y = 240 from the graph. Therefore, at 2 seconds, the ball is 240 feet high.

(b) At y = 192, x = 1 and x = 5 from the graph. Therefore, the height will be 192 feet at 1 second and at 5 seconds.

(c) The ball is going up from 0 to 3 seconds and down from 3 to 7 seconds.

(d) The coordinate of the highest point is (3, 256). Therefore, it reaches a maximum height of 256 feet at 3 seconds.

(e) At x = 7, y = 0. Therefore, at 7 seconds, the ball hits the ground.

95. (a) At t = 12 and t = 20, y = 55 from the graph. Therefore, after about 12 noon until about 8 P.M. the temperature was over 55º.

(b) At t = 6 and t = 22, y = 40 from the graph. Therefore, until about 6 A.M. and after 10 P.M. the temperature was below 40º.

(c) The temperature at noon in Bratenahl, Ohio was 55º. Because the temperature in Greenville is 7º higher, we are looking for the time at which Bratenahl, Ohio was at or above 48º. This occurred at approximately 10 A.M and 8:30 P.M.

(d) The temperature is just below 40° from midnight to 6 A.M., when it begins to rise until it reaches a maximum of just below 65° at 4 P.M. It then begins to fall util it reaches just under 40° again at midnight.

96. (a) At t = 8, y = 24 from the graph. Therefore, there are 24 units of the drug in the bloodstream at 8 hours.

(b) The level increases between 0 and 2 hours after the drug is taken and decreases between 2 and 12 hours after the drug is taken.

(c) The coordinates of the highest point are (2, 64). Therefore, at 2 hours, the level of the drug in the bloodstream reaches its greatest value of 64 units.

(d) After the peak, y = 16 at t = 10. 10 hours – 2 hours = 8 hours after the peak. 8 additional hours are required for the level to drop to 16 units.

Section 2.4 Linear Functions 195


(e) When the drug is administered, the level is 0 units. The level begins to rise quickly for 2 hours until it reaches a maximum of 64 units. The level then begins to decrease gradually until it reaches a level of 12 units, 12 hours after it was administered.

Section 2.4 Linear Functions

1. B; ( )f x = 3x + 6 is a linear function with y-intercept (0, 6).

2. H; x = 9 is a vertical line.

3. C; ( )f x = −8 is a constant function.

4. G; 2x – y = –4 or y = 2x + 4 is a linear equation with x-intercept (–2, 0) and y-intercept (0, 4).

5. A; ( )f x = 5x is a linear function whose graph passes through the origin, (0, 0). f(0) = 5(0) = 0.

6. D; 2( )f x x is a function that is not linear.

7. –3m matches graph C because the line falls rapidly as x increases.

8. m = 0 matches graph A because horizontal lines have slopes of 0.

9. m = 3 matches graph D because the line rises rapidly as x increases.

10. m is undefined for graph B because vertical lines have undefined slopes.

11. ( ) – 4f x x Use the intercepts.

(0) 0 – 4 – 4 :f y-intercept

0 – 4 4 :x x x-intercept Graph the line through (0, –4) and (4, 0).

The domain and range are both , .


(0) –0 4 4 :f y-intercept

0 – 4 4 :x x x-intercept Graph the line through (0, 4) and (4, 0).


13. 12

( ) – 6f x x

Use the intercepts.

12

(0) 0 – 6 – 6 :f y-intercept

1 12 2

0 – 6 6 12 :x x x x-intercept

Graph the line through (0, –6) and (12, 0).


14. 23

( ) 2f x x

Use the intercepts.

23

(0) 0 2 2 :f y-intercept

2 23 3

0 2 –2 3 : -interceptx x x x

Graph the line through (0, 2) and (–3, 0).




15. ( ) 3f x x The x-intercept and the y-intercept are both zero. This gives us only one point, (0, 0). If x = 1, y = 3(1) = 3 . Another point is (1, 3). Graph the line through (0, 0) and (1, 3).


16. –2f x x

The x-intercept and the y-intercept are both zero. This gives us only one point, (0, 0). If x = 3, y = –2(3) = –6, so another point is (3, –6). Graph the line through (0, 0) and (3, –6).


17. ( ) – 4f x is a constant function.

The graph of ( ) 4f x is a horizontal line with a y-intercept of –4.

domain: , ; range: {–4}

18. ( ) 3f x is a constant function whose graph is a horizontal line with y-intercept of 3.

domain: , ; range: {3}

19. ( ) 0f x is a constant function whose graph is the x-axis.


20. 9f x x


21. 4 3 12x y Use the intercepts.

4 0 3 12 3 124 : -intercept

y yy y

4 3 0 12 4 123 : -intercept

x xx x

Graph the line through (0, 4) and (−3, 0).


22. 2 5 10;x y Use the intercepts.

2 0 5 10 5 102 : -intercept

y yy y

2 5 0 10 2 105 : -intercept

x xx x

Graph the line through (0, 2) and (5, 0):




23. 3 4 0y x Use the intercepts.

3 4 0 0 3 0 0 : -intercepty y y y

3 0 4 0 4 0 0 : -interceptx x x x

The graph has just one intercept. Choose an additional value, say 3, for x.

3 4 3 0 3 12 03 12 4

y yy y



24. 3 2 0x y Use the intercepts. 3 0 2 0 2 0 0 : -intercepty y y y

3 2 0 0 3 0 0 :x x x x-intercept


3 2 2 0 6 2 02 6 3

y yy y

Graph the line through (0, 0) and (2, −3):


25. x = 3 is a vertical line, intersecting the x-axis at (3, 0).

domain: {3}; range: ,

26. x = –4 is a vertical line intersecting the x-axis at (–4, 0).

domain: {−4}; range: ,

27. 2x + 4 = 0 2 4 2x x is a vertical line intersecting the x-axis at (–2, 0).


28. 3 6 0 –3 6 2x x x is a vertical

line intersecting the x-axis at (2, 0).


29. 5 0 5x x is a vertical line intersecting the x-axis at (5, 0).




30. 3 0 3x x is a vertical line

intersecting the x-axis at 3,0 .


31. y = 5 is a horizontal line with y-intercept 5. Choice A resembles this.

32. y = –5 is a horizontal line with y-intercept –5. Choice C resembles this.

33. x = 5 is a vertical line with x-intercept 5. Choice D resembles this.

34. x = –5 is a vertical line with x-intercept –5. Choice B resembles this.

35.

36.

37.

38.

39. The rise is 2.5 feet while the run is 10 feet so

the slope is 2.5 110 4

0.25 25% . So A =

0.25, 2.510

C , D = 25%, and 14

E are all

expressions of the slope.

40. The pitch or slope is 14

. If the rise is 4 feet

then 1 rise 4

4 run x or x = 16 feet. So 16 feet in

the horizontal direction corresponds to a rise of 4 feet.

41. Through (2, –1) and (–3, –3) Let 1 1 22, –1, –3, and x y x 2 –3.y

Then rise –3 – (–1) –2y and

run –3 – 2 –5.x

The slope is rise –2 2

.run –5 5

ymx

42. Through (−3, 4) and (2, −8) Let 1 1 2 23, 4, 2, and –8.x y x y

Then rise –8 – 4 –12y and

run 2 – ( 3) 5.x


.run 5 5

ymx

43. Through (5, 8) and (3, 12) Let 1 1 2 25, 8, 3, and 12.x y x y

Then rise 12 8 4y and

run 3 5 2.x

The slope is rise 4

2.run 2

ymx

44. Through (5, –3) and (1, –7)

1 1 2 2Let 5, –3, 1, and –7.x y x y

Then rise –7 – (–3) – 4y and

run 1 – 5 – 4.x

The slope is 4

1.4

ymx

45. Through (5, 9) and (–2, 9)

2 1

2 1

9 – 9 00

–2 5 –7

y yymx x x

46. Through (–2, 4) and (6, 4)

2 1

2 1

4 4 00

6 (–2) 8

y yymx x x

47. Horizontal, through (5, 1) The slope of every horizontal line is zero, so m = 0.

48. Horizontal, through (3 , 5) The slope of every horizontal line is zero, so m = 0.

49. Vertical, through (4, –7) The slope of every vertical line is undefined; m is undefined.



50. Vertical, through (–8, 5) The slope of every vertical line is undefined; m is undefined.

51. (a) 3 5y x Find two ordered pairs that are solutions to the equation. If x = 0, then 3 0 5 5.y y

If x = −1, then 3 1 5 3 5 2.y y y

Thus two ordered pairs are (0, 5) and (−1, 2)

2 1

2 1

rise 2 5 33.

run 1 0 1

y ymx x

(b)

52. 2 4y x Find two ordered pairs that are solutions to the equation. If 0,x then 2 0 4y

4.y If 1,x then 2 1 4y

2 4 2.y y Thus two ordered pairs

are 0, 4 and 1, 2 .

2 1

2 1

2 4rise 22.

run 1 0 1

y ymx x

(b)

53. 2 3y x Find two ordered pairs that are solutions to the equation. If 0,x then 2 0 0.y y

If 3,y then 2 3 3 6 3x x

2.x Thus two ordered pairs are 0,0 and

2, 3 .

2 1

2 1

rise 3 0 3.

run 2 0 2

y ymx x

(b)


If 4,x then 4 5 4 4 20y y

5.y Thus two ordered pairs are 0,0 and

4, 5 .

2 1

2 1

rise 5 0 5.

run 4 0 4

y ymx x

(b)

55. 5 2 10x y Find two ordered pairs that are solutions to the equation. If 0,x then 5 0 2 10y

5.y If 0,y then 5 2 0 10x

5 10x 2.x

Thus two ordered pairs are 0, 5 and 2,0 .

2 1

2 1

0 5rise 5.

run 2 0 2

y ymx x

(b)




3 12y 4.y If 0,y then

4 3 0 12 4 12x x 3.x Thus two

ordered pairs are 0, 4 and 3,0 .

2 1

2 1

rise 0 4 4.

run 3 0 3

y ymx x

(b)

57. Through (–1, 3), 32

m

First locate the point (–1, 3). Because the

slope is 32

, a change of 2 units horizontally (2

units to the right) produces a change of 3 units vertically (3 units up). This gives a second point, (1, 6), which can be used to complete the graph.

58. Through (–2, 8), 25

m . Because the slope is

25

, a change of 5 units horizontally (to the

right) produces a change of 2 units vertically (2 units up). This gives a second point (3, 10), which can be used to complete the graph. Alternatively, a change of 5 units to the left produces a change of 2 units down. This gives the point (−7, 6).

59. Through (3, –4), 13

–m . First locate the point

(3, –4). Because the slope is 13

– , a change of

3 units horizontally (3 units to the right) produces a change of –1 unit vertically (1 unit down). This gives a second point, (6, –5), which can be used to complete the graph.

60. Through (–2, –3), 34

–m . Because the slope

is 3 –34 4

– , a change of 4 units horizontally

(4 units to the right) produces a change of –3 units vertically (3 units down). This gives a second point (2, –6), which can be used to complete the graph.

61. Through 12

, 4 , m = 0.

The graph is the horizontal line through

12

, 4 .

Exercise 61 Exercise 62

62. Through 32

, 2 , m = 0.


32

, 2 .



63. Through 52

, 3 , undefined slope. The slope

is undefined, so the line is vertical,

intersecting the x-axis at 52

, 0 .

64. Through 94

, 2 , undefined slope. The slope is

undefined, so the line is vertical, intersecting

the x-axis at 94

, 0 .

65. The average rate of change is f b f a

mb a

20 4 16$4

0 4 4

(thousand) per year. The

value of the machine is decreasing $4000 each year during these years.


mb a

200 0 200$50

4 0 4

per month. The amount

saved is increasing $50 each month during these months.

67. The graph is a horizontal line, so the average rate of change (slope) is 0. The percent of pay raise is not changing—it is 3% each year.

68. The graph is a horizontal line, so the average rate of change (slope) is 0. That means that the number of named hurricanes remained the same, 10, for the four consecutive years shown.

69. 2562 5085 2523

2012 1980 3278.8 thousand per year

f b f am

b a

The number of high school dropouts decreased by an average of 78.8 thousand per year from 1980 to 2012.

70. 1709 5302

2013 20063593

$513.297

f b f am

b a

Sales of plasma flat-panel TVs decreased by an average of $513.29 million per year from 2006 to 2013.

71. (a) The slope of –0.0167 indicates that the average rate of change of the winning time for the 5000 m run is 0.0167 min less. It is negative because the times are generally decreasing as time progresses.

(b) The Olympics were not held during World Wars I (1914−1919) and II (1939–1945).

(c) 0.0167 2000 46.45 13.05 miny

The model predicts a winning time of 13.05 minutes. The times differ by 13.35 − 13.05 = 0.30 min.

72. (a) From the equation, the slope is 200.02. This means that the number of radio stations increased by an average of 200.02 per year.

(b) The year 2018 is represented by x = 68. 200.02 68 2727.7 16,329.06y

According to the model, there will be about 16,329 radio stations in 2018.

73. 2013 2008 335,652 270,334

2013 2008 2013 200865,318

13,063.65

f f

The average annual rate of change from 2008 through 2013 is about 13,064 thousand.

74. 2014 2006 3.74 4.53

2014 2006 2014 20060.79

0.0998

f f

The average annual rate of change from 2006 through 2014 is about –0.099.



75. (a) 56.3 138

2013 200381.7

8.1710

f b f am

b a

The average rate of change was −8.17 thousand mobile homes per year.

(b) The negative slope means that the number of mobile homes decreased by an average of 8.17 thousand each year from 2003 to 2013.

76. 2013 1991 26.6 61.8

2013 1991 2013 199135.2

1.622

f f

There was an average decrease of 1.6 births per thousand per year from 1991 through 2013.

77. (a) 10 500C x x

(b) 35R x x

(c) 35 10 500

35 10 500 25 500

P x R x C xx xx x x

(d) 10 500 35

500 2520

C x R xx x

xx

20 units; do not produce

78. (a) 150 2700C x x

(b) 280R x x

(c) 280 150 2700

280 150 2700130 2700

P x R x C xx xx xx

(d) 150 2700 280

2700 13020.77 or 21 units

C x R xx x

xx

21 units; produce

79. (a) 400 1650C x x

(b) 305R x x

(c) 305 400 1650

305 400 165095 1650

P x R x C xx xx xx

(d) 400 1650 305

95 1650 095 1650

17.37 units

C x R xx xx

xx

This result indicates a negative “break-even point,” but the number of units produced must be a positive number. A calculator graph of the lines

1 400 1650y C x x and

2 305y R x x in the window

[0, 70] × [0, 20000] or solving the inequality 305 400 1650x x will show

that R x C x for all positive values

of x (in fact whenever x is greater than −17.4). Do not produce the product because it is impossible to make a profit.

80. (a) 11 180C x x

(b) 20R x x

(c) 20 11 180

20 11 180 9 180

P x R x C xx xx x x

(d) 11 180 20

180 920

C x R xx x

xx

20 units; produce

81. 200 1000 2401000 40 25C x R x x x

x x

The break-even point is 25 units. (25) 200 25 1000 $6000C which is the

same as (25) 240 25 $6000R

C x

R x

Chapter 2 Quiz (Sections 2.1–2.4) 203


82. 220 1000 2401000 20 50C x R x x x

x x

The break-even point is 50 units instead of 25 units. The manager is not better off because twice as many units must be sold before beginning to show a profit.

83. The first two points are A(0, –6) and B(1, –3). 3 – (–6) 3

31 – 0 1

m

84. The second and third points are B(1, –3) and C(2, 0).

0 – (–3) 33

2 – 1 1m

85. If we use any two points on a line to find its slope, we find that the slope is the same in all cases.

86. The first two points are A(0, –6) and B(1, –3). 2 2

2 2

( , ) (1 – 0) [–3 – (– 6)]

1 3 1 9 10

d A B

87. The second and fourth points are B(1, –3) and D(3, 3).

2 2

2 2

( , ) (3 – 1) [3 – (–3)]

2 6 4 36

40 2 10

d B D

88. The first and fourth points are A(0, –6) and D(3, 3).

2 2

2 2

( , ) (3 – 0) [3 – (– 6)]

3 9 9 81

90 3 10

d A D

89. 10 2 10 3 10; The sum is 3 10, which is equal to the answer in Exercise 88.

90. If points A, B, and C lie on a line in that order, then the distance between A and B added to the distance between B and C is equal to the distance between A and C.

91. The midpoint of the segment joining A(0, –6) and G(6, 12) has coordinates

0 6 6 122 2

, 6 62 2

, 3,3 . The midpoint is

M(3, 3), which is the same as the middle entry in the table.

92. The midpoint of the segment joining E(4, 6) and F(5, 9) has coordinates

4 5 6 9 9 152 2 2 2

, , = (4.5, 7.5). If the

x-value 4.5 were in the table, the corresponding y-value would be 7.5.

Chapter 2 Quiz (Sections 2.1−2.4)

1.

2 22 1 2 1

2 2

2 2

( , )

8 ( 4) 3 2

( 4) ( 5) 16 25 41

d A B x x y y

2. To find an estimate for 2006, find the midpoint of (2004, 6.55) and (2008, 6.97:

2004 2008 6.55 6.97,

2 22006, 6.76

M

The estimated enrollment for 2006 was 6.76 million. To find an estimate for 2010, find the midpoint of (2008, 6.97) and (2012, 7.50):

2008 2012 6.97 7.50,

2 22010, 7.235

M

The estimated enrollment for 2006 was about 7.24 million.

3.

4.


2 2

2 2

( 4 4) ( 8 16) 3 4 16

( 2) ( 4) 17

x x y yx y

The radius is 17 and the midpoint of the circle is (2, −4).

6. From the graph, f(−1) is 2.

7. Domain: ( , ); range: [0, )



8. (a) The largest open interval over which f is decreasing is ( , 3) .

(b) The largest open interval over which f is increasing is 3, .

(c) There is no interval over which the function is constant.

9. (a) 11 5 6 3

5 1 4 2m

(b) 4 4 0

01 ( 7) 6

m

(c) 4 12 16

6 6 0m

the slope is

undefined.

10. The points to use are (2009, 10,602) and (2013, 15,884). The average rate of change is

15,884 10,602 5282

1320.52013 2009 4

The average rate of change was 1320.5 thousand cars per year. This means that the number of new motor vehicles sold in the United States increased by an average of 1320.5 thousand per year from 2009 to 2013.

Section 2.5 Equations of Lines and Linear Models

1. The graph of the line 3 4 8y x has

slope 4 and passes through the point (8, 3).

2. The graph of the line 2 7y x has slope –2 and y-intercept (0, 7).

3. The vertical line through the point (–4, 8) has equation x = –4.

4. The horizontal line through the point (–4, 8) has equation y = 8.

For exercises 5 and 6, 67

6 7 0 7 6x y y x y x

5. Any line parallel to the graph of 6 7 0x y

must have slope 67.

6. Any line perpendicular to the graph of

6 7 0x y must have slope 76.

7. 14

2y x is graphed in D.

The slope is 14

and the y-intercept is (0, 2).

8. 4x + 3y = 12 or 3y = –4x + 12 43

or – 4y x

is graphed in B. The slope is 43

– and the

y-intercept is (0, 4).

9. 32

– 1 1y x is graphed in C. The slope

is 32

and a point on the graph is (1, –1).

10. y = 4 is graphed in A. y = 4 is a horizontal line with y-intercept (0, 4).

11. Through (1, 3), m = –2. Write the equation in point-slope form.

1 1– – 3 –2 –1y y m x x y x

Then, change to standard form. 3 –2 2 2 5y x x y

12. Through (2, 4), m = –1 Write the equation in point-slope form.

1 1– – 4 –1 – 2y y m x x y x

Then, change to standard form. – 4 – 2 6y x x y

13. Through (–5, 4), 32

m

Write the equation in point-slope form.

32

– 4 5y x

Change to standard form. 2 – 4 –3 52 – 8 –3 – 15

3 2 –7

y xy x

x y

14. Through 34

(– 4, 3), m


34

– 3 4y x

Change to standard form. 4 3 3 4

4 12 3 123 4 24 or 3 4 24

y xy xx y x y

15. Through (–8, 4), undefined slope Because undefined slope indicates a vertical line, the equation will have the form x = a. The equation of the line is x = –8.

16. Through (5, 1), undefined slope This is a vertical line through (5, 1), so the equation is x = 5.

17. Through (5, −8), m = 0 This is a horizontal line through (5, −8), so the equation is y = −8.

18. Through (−3, 12), m = 0 This is a horizontal line through (−3, 12), so the equation is y = 12.

Section 2.5 Equations of Lines and Linear Models 205


19. Through (–1, 3) and (3, 4) First find m.

4 – 3 1

3 – (–1) 4m

Use either point and the point-slope form.

14

– 4 – 34 16 3

4 134 13

y xy xx yx y

20. Through (2, 3) and (−1, 2) First find m.

2 3 1 1

1 2 3 3m


13

3 – 23 – 9 2

3 73 7

y xy x

x yx y

21. x-intercept (3, 0), y-intercept (0, –2) The line passes through (3, 0) and (0, –2). Use these points to find m.

–2 – 0 2

0 – 3 3m

Using slope-intercept form we have 23

– 2.y x

22. x-intercept (–4, 0), y-intercept (0, 3) The line passes through the points (–4, 0) and (0, 3). Use these points to find m.

3 – 0 3

0 – (–4) 4m


3.y x

23. Vertical, through (–6, 4) The equation of a vertical line has an equation of the form x = a. Because the line passes through (–6, 4), the equation is x = –6. (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)

24. Vertical, through (2, 7) The equation of a vertical line has an equation of the form x = a. Because the line passes through (2, 7), the equation is x = 2. (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)

25. Horizontal, through (−7, 4) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (−7, 4), the equation is y = 4.

26. Horizontal, through (−8, −2) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (−8, −2), the equation is y = −2.

27. m = 5, b = 15 Using slope-intercept form, we have

5 15.y x

28. m = −2, b = 12 Using slope-intercept form, we have

2 12.y x

29. Through (−2, 5), slope = −4

5 4 2

5 4 25 4 8

4 3

y xy xy x

y x

30. Through (4, −7), slope = −2 7 2 4

7 2 82 1

y xy x

y x

31. slope 0, y-intercept 32

0,

These represent 32

0 and .m b

Using slope-intercept form, we have

3 32 2

0 .y x y


0,

These represent 54

0 and .m b


5 54 4

0 .y x y

33. The line x + 2 = 0 has x-intercept (–2, 0). It does not have a y-intercept. The slope of his line is undefined. The line 4y = 2 has

y-intercept 12

0, . It does not have an

x-intercept. The slope of this line is 0.

34. (a) The graph of y = 3x + 2 has a positive slope and a positive y-intercept. These conditions match graph D.



(b) The graph of y = –3x + 2 has a negative slope and a positive y-intercept. These conditions match graph B.

(c) The graph of y = 3x – 2 has a positive slope and a negative y-intercept. These conditions match graph A.

(d) The graph of y = –3x – 2 has a negative slope and a negative y-intercept. These conditions match graph C.

35. y = 3x – 1 This equation is in the slope-intercept form, y = mx + b. slope: 3; y-intercept: (0, –1)

36. y = –2x + 7 slope: –2; y-intercept: (0, 7)

37. 4x – y = 7 Solve for y to write the equation in slope-intercept form. – – 4 7 4 – 7y x y x slope: 4; y-intercept: (0, –7)

38. 2x + 3y = 16 Solve the equation for y to write the equation in slope-intercept form.

1623 3

3 –2 16 –y x y x

slope: 23

– ; y-intercept: 163

0,

39. 3 34 4

4 –3 – or 0y x y x y x

slope: 34

; y-intercept (0, 0)

40. 1 12 2

2 or 0y x y x y x

slope is 12

; y-intercept: (0, 0)

41. 2 4x y Solve the equation for y to write the equation in slope-intercept form.

12

2 – 4 – 2y x y x

slope: 12

– ; y-intercept: (0, −2)




13

3 – 9 – 3y x y x

slope: 13

– ; y-intercept: (0,−3)

43. 32

1 0y x

Solve the equation for y to write the equation in slope-intercept form.

3 32 2

1 0 1y x y x

slope: 32


44. (a) Use the first two points in the table, A(–2, –11) and B(–1, –8).

–8 – (–11) 33

–1 – (–2) 1m

(b) When x = 0, y = −5. The y-intercept is (0, –5).

(c) Substitute 3 for m and –5 for b in the slope-intercept form.

3 – 5y mx b y x

45. (a) The line falls 2 units each time the x value increases by 1 unit. Therefore the slope is −2. The graph intersects the y-axis at the point (0, 1) and intersects the

x-axis at 12

, 0 , so the y-intercept is

(0, 1) and the x-intercept is 12

, 0 .

(b) An equation defining f is y = −2x + 1.

46. (a) The line rises 2 units each time the x value increases by 1 unit. Therefore the slope is 2. The graph intersects the y-axis at the point (0, −1) and intersects the

x-axis at 12


(0, −1) and the x-intercept is 12

, 0 .

(b) An equation defining f is y = 2x − 1.

47. (a) The line falls 1 unit each time the x value increases by 3 units. Therefore the slope

is 13

. The graph intersects the y-axis at

the point (0, 2), so the y-intercept is (0, 2). The graph passes through (3, 1) and will fall 1 unit when the x value increases by 3, so the x-intercept is (6, 0).

(b) An equation defining f is 13

2.y x

48. (a) The line rises 3 units each time the x value increases by 4 units. Therefore the

slope is 34

. The graph intersects the

y-axis at the point (0, −3) and intersects the x-axis at (4, 0), so the y-intercept is (0, −3) and the x-intercept is 4.


3.y x

49. (a) The line falls 200 units each time the x value increases by 1 unit. Therefore the slope is −200. The graph intersects the y-axis at the point (0, 300) and intersects

the x-axis at 32



, 0 .


50. (a) The line rises 100 units each time the x value increases by 5 units. Therefore the slope is 20. The graph intersects the y-axis at the point (0, −50) and intersects

the x-axis at 52



, 0 .




51. (a) through (–1, 4), parallel to x + 3y = 5 Find the slope of the line x + 3y = 5 by writing this equation in slope-intercept form.

513 3

3 5 3 – 5–

x y y xy x

The slope is 13

.

Because the lines are parallel, 13

is also

the slope of the line whose equation is to

be found. Substitute 13

m , 1 –1,x

and 1 4y into the point-slope form.

1 11313

– –

4 – –1

– 4 13 –12 – –1 3 11

y y m x xy xy x

y x x y

(b) Solve for y. 1 113 3

3 – 11 –y x y x

52. (a) through (3, –2), parallel to 2x – y = 5 Find the slope of the line 2x – y = 5 by writing this equation in slope-intercept form. 2 – 5 – –2 5

2 – 5x y y x

y x

The slope is 2. Because the lines are parallel, the slope of the line whose equation is to be found is also 2. Substitute m = 2, 1 3x , and 1 2y into the point-slope form.

1 1 – –

2 2 – 3 2 2 – 6–2 –8 or 2 – 8

y y m x xy x y x

x y x y

(b) Solve for y. y = 2x – 8

53. (a) through (1, 6), perpendicular to 3x + 5y = 1 Find the slope of the line 3x + 5y = 1 by writing this equation in slope-intercept form.

3 15 5

3 5 1 5 –3 1–

x y y xy x

This line has a slope of 35

. The slope of

any line perpendicular to this line is 53

,

because 3 55 3

1. Substitute 53

m ,

1 1,x and 1 6y into the point-slope form.

53

– 6 ( –1)3( – 6) 5( –1)3 – 18 5 – 5

–13 5 – 3 or 5 – 3 –13

y xy xy x

x y x y


3 5 13y x y x

54. (a) through (–2, 0), perpendicular to 8x – 3y = 7 Find the slope of the line 8x – 3y = 7 by writing the equation in slope-intercept form.

8 73 3

8 – 3 7 –3 –8 7–

x y y xy x


. The slope of


,

because 8 33 8

1.

Substitute 38

m , 1 2,x and 1 0y

into the point-slope form. 38

– 0 – ( 2)8 –3( 2)8 –3 – 6 3 8 –6

y xy xy x x y


3 38 4

8 –3 – 6 –

–

y x y xy x

55. (a) through (4, 1), parallel to y = −5 Because y = −5 is a horizontal line, any line parallel to this line will be horizontal and have an equation of the form y = b. Because the line passes through (4, 1), the equation is y = 1.

(b) The slope-intercept form is y = 1.

56. (a) through 2, 2 , parallel to y = 3.

Because y = 3 is a horizontal line, any line parallel to this line will be horizontal and have an equation of the form y = b. Because the line passes through 2, 2 ,

the equation is y = –2.

(b) The slope-intercept form is y = −2

57. (a) through (–5, 6), perpendicular to x = –2. Because x = –2 is a vertical line, any line perpendicular to this line will be horizontal and have an equation of the form y = b. Because the line passes through (–5, 6), the equation is y = 6.




58. (a) Through (4, –4), perpendicular to x = 4 Because x = 4 is a vertical line, any line perpendicular to this line will be horizontal and have an equation of the form y = b. Because the line passes through (4, –4), the equation is y = –4.

(b) The slope-intercept form is y = –4.

59. (a) Find the slope of the line 3y + 2x = 6.

23

3 2 6 3 –2 6– 2

y x y xy x

Thus, 23

– .m A line parallel to

3y + 2x = 6 also has slope 23

– .

Solve for k using the slope formula.

12

2 1 2–

4 33 2

–4 3

3 23 4 3 4 –

4 39 2 49 2 8

2 1

k

k

k kk

kk

k k

(b) Find the slope of the line 2y – 5x = 1.

5 12 2

2 5 1 2 5 1y x y xy x

Thus, 52

.m A line perpendicular to 2y

– 5x = 1 will have slope 25

– , because

5 22 5

– –1.

Solve this equation for k.

72

3 2

4 53 2

5 4 5 44 515 2 415 2 82 7

k

k kk

kk

k k

60. (a) Find the slope of the line 2 – 3 4.x y

2 43 3

2 – 3 4 –3 –2 4–

x y y xy x

Thus, 23

.m A line parallel to

2x – 3y = 4 also has slope 23

. Solve for r

using the slope formula. – 6 2 – 6 2

– 4 – 2 3 – 6 3

– 6 26 6

– 6 36 4 2

r r

r

r r

(b) Find the slope of the line x + 2y = 1.

1 12 2

2 1 2 – 1–

x y y xy x

Thus, 12

.m A line perpendicular to

the line x + 2y = 1 has slope 2, because 12

– (2) –1. Solve for r using the slope

formula. – 6 – 6

2 2– 4 – 2 – 6

– 6 –12 – 6

r r

r r

61. (a) First find the slope using the points (0, 6312) and (3, 7703).

7703 6312 1391463.67

3 0 3m

The y-intercept is (0, 6312), so the equation of the line is

463.67 6312.y x

(b) The value x = 4 corresponds to the year 2013.

463.67 4 6312 8166.68y

The model predicts that average tuition and fees were $8166.68 in 2013. This is $96.68 more than the actual amount.


7136 6312 824412

2 0 2m

The y-intercept is (0, 6312), so the equation of the line is 412 6312.y x


412 4 6312 7960y

The model predicts that average tuition and fees were $7960 in 2013. This is $110 less than the actual amount.




24525 22036 2489622.25

4 0 4m


622.25 22,036.y x

The slope of the line indicates that the

average tuition increase is about $622 per year from 2009 through 2013.

(b) The year 2012 corresponds to x = 3. 622.25 3 22,036 23,902.75y

According to the model, average tuition and fees were $23,903 in 2012. This is $443 more than the actual amount $23,460.

(c) Using the linear regression feature, the equation of the line of best fit is

653 21,634.y x

64. (a) See the graph in the answer to part (b).There appears to be a linear relationship between the data. The farther the galaxy is from Earth, the faster it is receding.

(b) Using the points (520, 40,000) and (0, 0), we obtain

40,000 – 0 40,00076.9.

520 – 0 520m

The equation of the line through these two points is y = 76.9x.

(c) 76.9 60,00060,000

78076.9

x

x x

According to the model, the galaxy Hydra is approximately 780 megaparsecs away.

(d) 11

1110 9

9.5 10

9.5 101.235 10 12.35 10

76.9

Am

A

Using m = 76.9, we estimate that the age of the universe is approximately 12.35 billion years.

(e) 11

10 9

119

9.5 101.9 10 or 19 10

509.5 10

9.5 10100

A

A

The range for the age of the universe is between 9.5 billion and 19 billion years.

65. (a) The ordered pairs are (0, 32) and (100, 212).

The slope is 212 – 32 180 9

.100 – 0 100 5

m

Use 91 1 5

( , ) (0,32) and x y m in the

point-slope form.

1 195959 95 5

– ( – )

– 32 ( – 0)

– 32

32 32

y y m x xy xy x

y x F C

(b)

95

59

32

5 9 325 9 160 9 5 –1609 5( – 32) ( – 32)

F CF CF C C FC F C F

(c) 59

( – 32)9 5( – 32) 9 5 – 1604 –160 – 40

F C F FF F F FF F

F = C when F is –40º.



66. (a) The ordered pairs are (0, 1) and (100, 3.92). The slope is

3.92 – 1 2.920.0292

100 – 0 100m and 1.b

Using slope-intercept form we have 0.0292 1y x or ( ) 0.0292 1.p x x

(b) Let x = 60. P(60) = 0.0292(60) + 1 = 2.752 The pressure at 60 feet is approximately 2.75 atmospheres.

67. (a) Because we want to find C as a function of I, use the points (12026, 10089) and (14167, 11484), where the first component represents the independent variable, I. First find the slope of the line.

11484 10089 13950.6516

14167 12026 2141m

Now use either point, say (12026, 10089), and the point-slope form to find the equation.

–10089 0.6516( –12026)–10089 0.6516 – 7836

0.6516 2253

C IC I

C I

(b) Because the slope is 0.6516, the marginal propensity to consume is 0.6516.

68. D is the only possible answer, because the x-intercept occurs when y = 0. We can see from the graph that the value of the x-intercept exceeds 10.

69. Write the equation as an equivalent equation with 0 on one side: 2 7 4 2x x x 2 7 4 2 0x x x . Now graph

2 7 4 2y x x x in the window [–5, 5] × [–5, 5] to find the x-intercept:

Solution set: {3}

70. Write the equation as an equivalent equation with 0 on one side: 7 2 4 5 3 1x x x 7 2 4 5 3 1 0x x x . Now graph

7 2 4 5 3 1y x x x in the window [–5, 5] × [–5, 5] to find the x-intercept:

Solution set: {1}

71. Write the equation as an equivalent equation with 0 on one side: 3 2 1 2 2 5x x

3 2 1 2 2 5 0x x . Now graph

3 2 1 2 2 5y x x in the window

[–5, 5] × [–5, 5] to find the x-intercept:

Solution set: 12

or 0.5

72. Write the equation as an equivalent equation with 0 on one side:

4 3 4 2 2 3 6 2x x x x

4 3 4 2 2 3 6 2 0x x x x .

Now graph 4 3 4 2 2 3 6 2y x x x x in the

window [–2, 8] × [–5, 5] to find the x-intercept:

Solution set: {4}

73. (a) 2 5 22 10 2

10 212

x xx x

xx

Solution set: {12}



(b) Answers will vary. Sample answer: The solution does not appear in the standard viewing window x-interval [10, –10]. The minimum and maximum values must include 12.

74. Rewrite the equation as an equivalent equation with 0 on one side.

3 2 6 4 8 2

6 18 4 8 2 0

x x xx x x

Now graph y = −6x − 18 − (−4x + 8 −2x) in the window [–10, 10] × [–30, 10].

The graph is a horizontal line that does not

intersect the x-axis. Therefore, the solution set is . We can verify this algebraically.

3 2 6 4 8 26 18 6 8 0 26

x x xx x

Because this is a false statement, the solution set is .

75. A(–1, 4), B(–2, –1), C(1, 14)

For A and B, 1 4 5

52 ( 1) 1

m

For B and C, 14 ( 1) 15

51 ( 2) 3

m

For A and C, 14 4 10

51 ( 1) 2

m

Since all three slopes are the same, the points are collinear.

76. A(0, −7), B(–3, 5), C(2, −15)

For A and B, 5 ( 7) 12

43 0 3

m

For B and C, 15 5 20

42 ( 3) 5

m

For A and C, 15 ( 7) 8

42 0 2

m


77. A(−1, −3), B(−5, 12), C(1, −11)

For A and B, 12 ( 3) 15

5 ( 1) 4m

For B and C, 11 12 23

1 ( 5) 6m

For A and C, 11 ( 3) 8

41 ( 1) 2

m

Since all three slopes are not the same, the points are not collinear.

78. A(0, 9), B(–3, –7), C(2, 19)

For A and B, 7 9 16 16

3 0 3 3m

For B and C, 19 ( 7) 26

2 ( 3) 5m


52 0 2

m

Because all three slopes are not the same, the points are not collinear.

79. 2 21 1 1

2 2 21 1 1

( , ) ( – 0) ( – 0)d O P x m x

x m x

80. 2 22 2 2

2 2 22 2 2

( , ) ( – 0) ( – 0)d O Q x m x

x m x

81. 2 22 1 2 2 1 1( , ) ( – ) ( – )d P Q x x m x m x

82.

2 2 2

2 22 2 2 2 2 2

1 1 1 2 2 2

22 2

2 1 2 2 1 1

2 2 2 2 2 21 1 1 2 2 2

2 22 1 2 2 1 1

[ ( , )] [ ( , )] [ ( , )]

– –

– –

d O P d O Q d P Q

x m x x m x

x x m x m x

x m x x m x

x x m x m x

2 2 2 2 2 21 1 1 2 2 2

2 2 2 22 2 1 1 2 2

2 21 2 1 2 1 1

2 1 1 2 1 2

1 2 1 2 1

2

20 2 2

2 2 0

x m x x m xx x x x m x

m m x x m xx x m m x x

m m x x x x

83. 1 2 1 2 1 2

1 2 1 2

–2 – 2 0–2 ( 1) 0m m x x x x

x x m m

84. 1 2 1 2–2 ( 1) 0x x m m

Because 1 20 and 0, x x we have

1 2 1 21 0 implying that –1.m m m m

Summary Exercises on Graphs, Circles, Functions, and Equations 213


85. If two nonvertical lines are perpendicular, then the product of the slopes of these lines is –1.

Summary Exercises on Graphs, Circles, Functions, and Equations

1. (3, 5), (2, 3)P Q

(a)

2 2

2 2

( , ) (2 3) ( 3 5)

1 8

1 64 65

d P Q


5 33 2 5 2 5, , ,1

2 2 2 2 2

.

(c) First find m: 3 – 5 8

82 – 3 1

m

Use either point and the point-slope form. – 5 8 – 3y x

Change to slope-intercept form. 5 8 24 8 19y x y x

2. P(–1, 0), Q(4, –2)

(a) 2 2

2 2

( , ) [4 – (–1)] (–2 – 0)

5 (–2)

25 4 29

d P Q


–1 4 0 (–2) 3 2, ,

2 2 2 23

, 1 .2

(c) First find m: 2 – 0 2 2

4 – 1 5 5m


25

– 0 – 1y x

Change to slope-intercept form.

2 25 5

5 2 15 2 2

y xy xy x

3. P(–2, 2), Q(3, 2)

(a) 22

2 2

( , ) [3 – (– 2)] 2 2

5 0 25 0 25 5

d P Q


– 2 3 2 2,

2 2

1 4 1, , 2 .

2 2 2


03 – 2 5

m

All lines that have a slope of 0 are horizontal lines. The equation of a horizontal line has an equation of the form y = b. Because the line passes through (3, 2), the equation is y = 2.

4. 2 2, 2 ,P 2,3 2Q

(a)

2 2

2 2

( , ) 2 – 2 2 3 2 – 2

– 2 2 2

2 8 10

d P Q


2 2 2 2 3 2,

2 2

3 2 4 2 3 2, , 2 2 .

2 2 2

(c) First find m: 3 2 2 2 2

22 2 2 2

m


– 2 2 – 2 2y x


– 2 2 4 2 2 5 2y x y x

5. P(5, –1), Q(5, 1)

(a) 2 2

2 2

( , ) (5 – 5) [1 – (–1)]

0 2 0 4 4 2

d P Q


5 5 –1 1,

2 2

10 0

, 5,0 .2 2



(c) First find m. 1 – 1 2

undefined5 – 5 0

m

All lines that have an undefined slope are vertical lines. The equation of a vertical line has an equation of the form x = a. The line passes through (5, 1), so the equation is x = 5 . (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)

6. (1, 1), ( 3, 3)P Q

(a)

2 2

2 2

( , ) ( 3 – 1) ( 3 –1)

4 4

16 16 32 4 2

d P Q


1 3 1 3,

2 2

2 2,

2 2

1, 1 .

(c) First find m: 3 –1 4

13 –1 4

m

Use either point and the point-slope form. –1 1 –1y x

Change to slope-intercept form. 1 1y x y x

7. 2 3,3 5 , 6 3,3 5P Q

(a)

2 2

2 2

( , ) 6 3 – 2 3 3 5 – 3 5

4 3 0 48 4 3

d P Q


2 3 6 3 3 5 3 5,

2 2

8 3 6 5, 4 3, 3 5 .

2 2


06 3 2 3 4 3

m

All lines that have a slope of 0 are horizontal lines. The equation of a horizontal line has an equation of the form y = b. Because the line passes

through 2 3,3 5 , the equation is

3 5.y

8. P(0, –4), Q(3, 1)

(a) 2 2

2 2

( , ) (3 – 0) [1 – (– 4)]

3 5 9 25 34

d P Q


0 3 – 4 1 3 3 3 3, , , .

2 2 2 2 2 2


3 0 3m


– 4.y x

9. Through 2,1 and 4, 1

First find m: 1 –1 2 1

4 – (–2) 6 3m


13

– 1 – 4y x


1 13 3

3 1 4 3 3 4

3 1

y x y xy x y x



10. the horizontal line through (2, 3) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (2, 3), the equation is y = 3.

11. the circle with center (2, –1) and radius 3

22 2

2 2

( – 2) – (–1) 3

( – 2) ( 1) 9

x yx y

12. the circle with center (0, 2) and tangent to the x-axis The distance from the center of the circle to the x-axis is 2, so r = 2.

2 2 2 2 2( – 0) ( – 2) 2 ( – 2) 4x y x y

13. the line through (3, 5) with slope 56


56

– 5 3 y x

Change to standard form.

5 156 6

5 56 2

6 5 5 3 6 30 5 15

6 5 15

y x y xy x y x

y x

14. the vertical line through (−4, 3) The equation of a vertical line has an equation of the form x = a. Because the line passes through (−4, 3), the equation is x = −4.

15. a line through (–3, 2) and parallel to the line 2x + 3y = 6 First, find the slope of the line 2x + 3y = 6 by writing this equation in slope-intercept form.

23

2 3 6 3 2 6 2x y y x y x

The slope is 23

. Because the lines are

parallel, 23

is also the slope of the line

whose equation is to be found. Substitute 23

m , 1 3x , and 1 2y into the point-

slope form.

21 1 3

23

– – 2 3

3 2 2 3 3 6 2 – 6

3 –2

y y m x x y xy x y x

y x y x

16. a line through the origin and perpendicular to the line 3 4 2x y

First, find the slope of the line 3 4 2x y by writing this equation in slope-intercept form.

3 32 14 4 4 2

3 4 2 4 –3 2x y y xy x y x


. The slope of any

line perpendicular to this line is 43

, because

343 4

1. Using slope-intercept form we

have 4 43 3

0 or .y x y x




(continued)

17. 2 24 2 4x x y y Complete the square on x and y separately.

2 2

2 2

2 2

4 2 4

4 4 2 1 4 4 1

2 1 9

x x y y

x x y y

x y

Yes, it is a circle. The circle has its center at (2, −1) and radius 3.

18. 2 26 10 36 0x x y y Complete the square on x and y separately.

2 2

2 2

2 2

6 10 36

6 9 10 25 –36 9 25

3 5 2

x x y y

x x y y

x y

No, it is not a circle.

19. 2 212 20 0x x y Complete the square on x and y separately.

2 2

2 2

2 2

12 20

12 36 –20 36

6 16

x x y

x x y

x y

Yes, it is a circle. The circle has its center at (6, 0) and radius 4.


2 2

2 2

2 2

2 16 61

2 1 16 64 –61 1 64

1 8 4

x x y y

x x y y

x y

Yes, it is a circle. The circle has its center at (−1, −8) and radius 2.


2 2

2 2

2 2

2 10

2 1 –10 1

1 9

x x y

x x y

x y


22. 2 2 – 8 9 0x y y Complete the square on x and y separately.

2 2

2 2

22

– 8 9

– 8 16 9 16

– 4 25

x y y

x y y

x y


23. The equation of the circle is 2 2 2( 4) ( 5) 4 .x y

Let y = 2 and solve for x: 2 2 2( 4) (2 5) 4x 2 2 2 2( 4) ( 3) 4 ( 4) 7x x

4 7 4 7x x

The points of intersection are 4 7, 2 and

4 7, 2

24. Write the equation in center-radius form by completing the square on x and y separately:

2 2

2 2

2 2

2 2

10 24 144 0

10 24 144 0

( 10 25) ( 24 144) 25

( 5) ( 12) 25

x y x yx x y y

x x y yx y

The center of the circle is (5, 12) and the radius is 5. Now use the distance formula to find the distance from the center (5, 12) to the origin:

2 22 1 2 1

2 2(5 0) (12 0) 25 144 13

d x x y y

The radius is 5, so the shortest distance from the origin to the graph of the circle is 13 − 5 = 8.


Section 2.6 Graphs of Basic Functions 217


(continued)

25. (a) The equation can be rewritten as 6 31 1

4 4 4 24 6 .y x y x y x

x can be any real number, so the domain is all real numbers and the range is also all real numbers. domain: , ; range: ,

(b) Each value of x corresponds to just one value of y. 4 6x y represents a function.

3 31 14 2 4 2

y x f x x

3 31 1 24 2 2 2 2

2 2 1f

26. (a) The equation can be rewritten as 2 5 .y x y can be any real number.

Because the square of any real number is

not negative, 2y is never negative. Taking the constant term into consideration, domain would be 5, .

domain: 5, ; range: ,

(b) Because (−4, 1) and (−4, −1) both satisfy

the relation, 2 5y x does not represent a function.

27. (a) 2 22 25x y is a circle centered at

(−2, 0) with a radius of 5. The domain will start 5 units to the left of –2 and end 5 units to the right of –2. The domain will be [−2 − 5, 2 + 5] = [−7, 3]. The range will start 5 units below 0 and end 5 units above 0. The range will be [0 − 5, 0 + 5] = [−5, 5].


the relation, 2 22 25x y does not

represent a function.


2 22 3 .y x y x x can be

any real number. Because the square of

any real number is not negative, 212

x is

never negative. Taking the constant term into consideration, range would be

32

, .

domain: , ; range: 32

,

(b) Each value of x corresponds to just one

value of y. 2 2 3x y represents a function.

2 23 31 12 2 2 2

y x f x x

2 3 3 31 1 4 12 2 2 2 2 2 2

2 2 4f

Section 2.6 Graphs of Basic Functions

1. The equation 2f x x matches graph E.

The domain is , .

2. The equation of f x x matches graph G.

The function is increasing on 0, .

3. The equation 3f x x matches graph A.

The range is , .

4. Graph C is not the graph of a function.

Its equation is 2x y .

5. Graph F is the graph of the identity function. Its equation is .f x x

6. The equation f x x matches graph B.

1.5 1f

7. The equation 3f x x matches graph H.

No, there is no interval over which the function is decreasing.

8. The equation of f x x matches graph D.

The domain is 0, .

9. The graph in B is discontinuous at many points. Assuming the graph continues, the range would be {…, −3, −2, −1, 0, 1, 2, 3, …}.



10. The graphs in E and G decrease over part of the domain and increase over part of the domain. They both increase over 0, and

decrease over , 0 .

11. The function is continuous over the entire domain of real numbers , .


13. The function is continuous over the interval

0, .

14. The function is continuous over the interval , 0 .

15. The function has a point of discontinuity at (3, 1). It is continuous over the interval ,3 and the interval 3, .

16. The function has a point of discontinuity at x = 1. It is continuous over the interval ,1

and the interval 1, .

17. 2 if –1( )

–1 if –1x xf x x x

(a) (–5) 2(–5) –10f

(b) (–1) 2(–1) –2f

(c) f(0) = 0 –1 = –1

(d) f(3) = 3 – 1 = 2

18. – 2 if 3( )

5 – if 3x xf x x x

(a) f(–5) = –5 – 2 = –7

(b) f(–1) = –1 – 2 = –3

(c) f(0) = 0 – 2 = –2

(d) f(3) = 5 – 3 = 2

19. 2 if – 4

( ) – if – 4 23 if 2

x xf x x x

x x

(a) f(–5) = 2 + (–5) = –3

(b) f(–1) = – (–1) = 1

(c) f(0) = –0 = 0

(d) (3) 3 3 9f

20. –2 if –33 –1 if – 3 2– 4 if 2

x xf x x x

x x

(a) f(–5) = –2(–5) = 10

(b) f(–1) = 3(–1) – 1 = –3 – 1 = –4

(c) f(0) = 3(0) – 1 = 0 – 1 = –1

(d) f(3) = –4(3) = –12

21. – 1 if 3( )

2 if 3x xf x x

Draw the graph of y = x – 1 to the left of x = 3, including the endpoint at x = 3. Draw the graph of y = 2 to the right of x = 3, and note that the endpoint at x = 3 coincides with the endpoint of the other ray.

22. 6 – if 3( )

3 if 3x xf x x

Graph the line y = 6 – x to the left of x = 3, including the endpoint. Draw y = 3 to the right of x = 3. Note that the endpoint at x = 3 coincides with the endpoint of the other ray.

23. 4 – if 2( )

1 2 if 2x xf x x x

Draw the graph of y = 4 − x to the left of x = 2, but do not include the endpoint. Draw the graph of y = 1 + 2x to the right of x = 2, including the endpoint.



24. 2 1 if 0( )

if 0x xf x x x

Graph the line y = 2x + 1 to the right of x = 0, including the endpoint. Draw y = x to the left of x = 0, but do not include the endpoint.

25. 3 if 1( )

1 if 1xf x x

Graph the line y = −3 to the left of x = 1, including the endpoint. Draw y = −1 to the right of x = 1, but do not include the endpoint.

26. 2 if 1( )

2 if 1xf x x

Graph the line y = – 2 to the left of x = 1, including the endpoint. Draw y = 2 to the right of x = 1, but do not include the endpoint.


27. 2 if – 4

( ) – if – 4 53 if 5

x xf x x x

x x

Draw the graph of y = 2 + x to the left of –4, but do not include the endpoint at x = 4. Draw the graph of y = –x between –4 and 5, including both endpoints. Draw the graph of y = 3x to the right of 5, but do not include the endpoint at x = 5.

28. –2 if –3

( ) 3 –1 if – 3 2– 4 if 2

x xf x x x

x x

Graph the line y = –2x to the left of x = –3, but do not include the endpoint. Draw y = 3x – 1 between x = –3 and x = 2, and include both endpoints. Draw y = –4x to the right of x = 2, but do not include the endpoint. Notice that the endpoints of the pieces do not coincide.

29. 21

212

– +2 if 2( )

if 2

x xf x

x x

Graph the curve 212

2y x to the left of

x = 2, including the endpoint at (2, 0). Graph

the line 12

y x to the right of x = 2, but do

not include the endpoint at (2, 1). Notice that the endpoints of the pieces do not coincide.

30. 3

25 if 0

( )if 0

x xf xx x

Graph the curve 3 5y x to the left of x = 0, including the endpoint at (0, 5). Graph

the line 2y x to the right of x = 0, but do not include the endpoint at (0, 0). Notice that the endpoints of the pieces do not coincide.



31. 2

2 if 5 1( ) 2 if 1 0

2 if 0 2

x xf x x

x x

Graph the line 2y x between x = −5 and x = −1, including the left endpoint at (−5, −10), but not including the right endpoint at (−1, −2). Graph the line y = −2 between x = −1 and x = 0, including the left endpoint at (−1, −2) and not including the right endpoint at (0, −2). Note that (−1, −2) coincides with the first two sections, so it is included. Graph

the curve 2 2y x from x = 0 to x = 2, including the endpoints at (0, −2) and (2, 2).

Note that (0, −2) coincides with the second two sections, so it is included. The graph ends at x = −5 and x = 2.

32.

2

2

0.5 if 4 2( ) if 2 2

4 if 2 4

x xf x x x

x x

Graph the curve 20.5y x between x = −4 and x = −2, including the endpoints at (−4, 8) and (−2, 2). Graph the line y x between x = −2 and x = 2, but do not include the endpoints at (−2, −2) and (2, 2). Graph the

curve 2 4y x from x = 2 to x = 4, including the endpoints at (2, 0) and (4, 12). The graph ends at x = −4 and x = 4.

33.

3

2

3 if 2 0( ) 3 if 0 1

4 if 1 3

x xf x x x

x x x

Graph the curve 3 3y x between x = −2 and x = 0, including the endpoints at (−2, −5) and (0, 3). Graph the line y = x + 3 between x = 0 and x = 1, but do not include the endpoints at (0, 3) and (1, 4). Graph the curve

24y x x from x = 1 to x = 3, including the endpoints at (1, 4) and (3, −2). The graph ends at x = −2 and x = 3.

34. 2

312

2 if 3 1

( ) 1 if 1 2

1 if 2 3

x xf x x x

x x

Graph the curve y = −2x to from x = −3 to x = −1, including the endpoint (−3, 6), but not including the endpoint (−1, 2). Graph the

curve 2 1y x from x = −1 to x = 2, including the endpoints (−1, 2) and (2, 5).

Graph the curve 312

1y x from x = 2 to

x = 3, including the endpoint (3, 14.5) but not including the endpoint (2, 5). Because the endpoints that are not included coincide with endpoints that are included, we use closed dots on the graph.



35. The solid circle on the graph shows that the endpoint (0, –1) is part of the graph, while the open circle shows that the endpoint (0, 1) is not part of the graph. The graph is made up of parts of two horizontal lines. The function which fits this graph is

–1 if 0( )

1 if 0.xf x x

domain: , ; range: {–1, 1}

36. We see that y = 1 for every value of x except x = 0, and that when x = 0, y = 0. We can write the function as

1 if 0( )

0 if 0.xf x x

domain: , ; range: {0, 1}

37. The graph is made up of parts of two horizontal lines. The solid circle shows that the endpoint (0, 2) of the one on the left belongs to the graph, while the open circle shows that the endpoint (0, –1) of the one on the right does not belong to the graph. The function that fits this graph is

2 if 0( )

–1 if 1.xf x x

domain: ( , 0] (1, ); range: {–1, 2}

38. We see that y = 1 when 1x and that y = –1 when x > 2. We can write the function as

1 if –1( )

–1 if 2.xf x x

domain: (– , –1] (2, ); range: {–1, 1}

39. For 0x , that piece of the graph goes through the points (−1, −1) and (0, 0). The slope is 1, so the equation of this piece is y = x. For x > 0, that piece of the graph is a horizontal line passing through (2, 2), so its equation is y = 2. We can write the function as

if 0( ) .

2 if 0x xf x x

domain: (– , ) range: ( , 0] {2}

40. For x < 0, that piece of the graph is a horizontal line passing though (−3, −3), so the equation of this piece is y = −3. For 0x , the curve passes through (1, 1) and (4, 2), so the

equation of this piece is y x . We can

write the function as 3 if 0

( ) . if 0

xf x

x x

domain: (– , ) range: { 3} [0, )

41. For x < 1, that piece of the graph is a curve passes through (−8, −2), (−1, −1) and (1, 1), so

the equation of this piece is 3y x . The right piece of the graph passes through (1, 2) and

(2, 3). 2 3

11 2

m

, and the equation of the

line is 2 1 1y x y x . We can write

the function as 3 if 1( )

1 if 1x xf x

x x

domain: (– , ) range: ( ,1) [2, )

42. For all values except x = 2, the graph is a line. It passes through (0, −3) and (1, −1). The slope is 2, so the equation is y = 2x −3. At x = 2, the graph is the point (2, 3). We can write

the function as 3 if 2( ) .

2 3 if 2xf x x x

domain: (– , ) range: ( ,1) (1, )

43. f(x) = x

Plot points.

x –x f(x) = x

–2 2 2

–1.5 1.5 1

–1 1 1

–0.5 0.5 0

0 0 0

0.5 –0.5 –1

1 –1 –1

1.5 –1.5 –2

2 –2 –2

More generally, to get y = 0, we need 0 – 1 0 1 1 0.x x x To get y = 1, we need 1 – 2x

1 2 –2 1.x x Follow this pattern to graph the step function.

domain: , ; range: {...,–2,–1,0,1,2,...}



44. f(x) = x

Plot points.

x x f(x) = x

–2 −2 2

–1.5 −2 2

–1 −1 1

–0.5 −1 1

0 0 0

0.5 0 0

1 1 −1

1.5 1 –1

2 2 –2

Follow this pattern to graph the step function.

domain: , ; range: {...,–2,–1,0,1,2,...}

45. f(x) = 2x

To get y = 0, we need 12

0 2 1 0 .x x


1 2 2 1.x x


2 2 3 1 .x x


domain: , ; range: {...,–2,–1,0,1,2,...}

46. g(x) = 2 1x


0 2 1 1 1 2 2 1.x x x


1 2 – 1 2 2 2 3 1 .x x x


domain: , ; range: {..., 2,–1,0,1,2,...}

47. The cost of mailing a letter that weighs more than 1 ounce and less than 2 ounces is the same as the cost of a 2-ounce letter, and the cost of mailing a letter that weighs more than 2 ounces and less than 3 ounces is the same as the cost of a 3-ounce letter, etc.

48. The cost is the same for all cars parking

between 12

hour and 1-hour, between 1 hour

and 12

1 hours, etc.

49.

Section 2.7 Graphing Techniques 223


50.

51. (a) For 0 8,x 49.8 34.2

1.958 0

m

,

so 1.95 34.2.y x For 8 13x ,

52.2 49.80.48

13 8m

, so the equation

is 52.2 0.48( 13)y x 0.48 45.96y x

(b) 1.95 34.2 if 0 8( )

0.48 45.96 if 8 13x xf x x x

52. When 0 3x , the slope is 5, which means that the inlet pipe is open, and the outlet pipe is closed. When 3 5x , the slope is 2, which means that both pipes are open. When 5 8x , the slope is 0, which means that both pipes are closed. When 8 10x , the slope is −3, which means that the inlet pipe is closed, and the outlet pipe is open.

53. (a) The initial amount is 50,000 gallons. The final amount is 30,000 gallons.

(b) The amount of water in the pool remained constant during the first and fourth days.

(c) (2) 45,000; (4) 40,000f f

(d) The slope of the segment between (1, 50000) and (3, 40000) is −5000, so the water was being drained at 5000 gallons per day.

54. (a) There were 20 gallons of gas in the tank at x = 3.

(b) The slope is steepest between t = 1 and t ≈ 2.9, so that is when the car burned gasoline at the fastest rate.

55. (a) There is no charge for additional length, so we use the greatest integer function. The cost is based on multiples of two

feet, so 2

( ) 0.8 xf x if 6 18x .

(b) 8.52

(8.5) 0.8 0.8(4) $3.20f

15.22

(15.2) 0.8 0.8(7) $5.60f

56. (a) 6.5 if 0 4

( ) 5.5 48 if 4 6–30 195 if 6 6.5

x xf x x x

x x

Draw a graph of y = 6.5x between 0 and 4, including the endpoints. Draw the graph of y = –5.5x + 48 between 4 and 6, including the endpoint at 6 but not the one at 4. Draw the graph of y = –30x + 195, including the endpoint at 6.5 but not the one at 6. Notice that the endpoints of the three pieces coincide.

(b) From the graph, observe that the snow depth, y, reaches its deepest level (26 in.) when x = 4, x = 4 represents 4 months after the beginning of October, which is the beginning of February.

(c) From the graph, the snow depth y is nonzero when x is between 0 and 6.5. Snow begins at the beginning of October and ends 6.5 months later, in the middle of April.

Section 2.7 Graphing Techniques

1. To graph the function 2 3,f x x shift the

graph of 2y x down 3 units.


graph of 2y x up 5 units.

3. The graph of 24f x x is obtained by

shifting the graph of 2y x to the left 4 units.


shifting the graph of 2y x to the right 7 units.

5. The graph of f x x is a reflection of

the graph of f x x across the x-axis.


the graph of f x x across the y-axis.



7. To obtain the graph of 32 3,f x x

shift the graph of 3y x 2 units to the left and 3 units down.


shift the graph of 3y x 3 units to the right and 6 units up.

9. The graph of f x x is the same as the

graph of y x because reflecting it across

the y-axis yields the same ordered pairs.

10. The graph of 2x y is the same as the graph

of 2x y because reflecting it across the

x-axis yields the same ordered pairs.

11. (a) B; 2( 7)y x is a shift of 2 ,y x 7 units to the right.

(b) D; 2 7y x is a shift of 2 ,y x 7 units downward.

(c) E; 27y x is a vertical stretch of 2 ,y x by a factor of 7.

(d) A; 2( 7)y x is a shift of 2 ,y x 7 units to the left.

(e) C; 2 7y x is a shift of 2 ,y x 7 units upward.

12. (a) E; 34y x is a vertical stretch

of 3 ,y x by a factor of 4.

(b) C; 3y x is a reflection of 3 ,y x over the x-axis.

(c) D; 3y x is a reflection of 3 ,y x over the y-axis.

(d) A; 3 4y x is a shift of 3 ,y x 4 units to the right.

(e) B; 3 4y x is a shift of 3 ,y x 4 units down.

13. (a) B; 2 2y x is a shift of 2 ,y x 2 units upward.

(b) A; 2 2y x is a shift of 2 ,y x 2 units downward.

(c) G; 2( 2)y x is a shift of 2 ,y x 2 units to the left.

(d) C; 2( 2)y x is a shift of 2 ,y x 2 units to the right.

(e) F; 22y x is a vertical stretch of 2 ,y x by a factor of 2.

(f) D; 2y x is a reflection of 2 ,y x across the x-axis.

(g) H; 2( 2) 1y x is a shift of 2 ,y x 2 units to the right and 1 unit upward.

(h) E; 2( 2) 1y x is a shift of 2 ,y x 2 units to the left and 1 unit upward.

(i) I; 2( 2) 1y x is a shift of 2 ,y x 2 units to the left and 1 unit down.

14. (a) G; 3y x is a shift of ,y x 3 units to the left.

(b) D; 3y x is a shift of ,y x 3 units downward.

(c) E; 3y x is a shift of ,y x 3 units upward.

(d) B; 3y x is a vertical stretch of

,y x by a factor of 3.

(e) C; y x is a reflection of y x across the x-axis.

(f) A; 3y x is a shift of ,y x 3 units to the right.

(g) H; 3 2y x is a shift of ,y x 3 units to the right and 2 units upward.

(h) F; 3 2y x is a shift of ,y x 3 units to the left and 2 units upward.

(i) I; 3 2y x is a shift of ,y x 3 units to the right and 2 units downward.

15. (a) F; 2y x is a shift of y x 2 units

to the right.

(b) C; 2y x is a shift of y x 2 units

downward.

(c) H; 2y x is a shift of y x 2 units

upward.



(d) D; 2y x is a vertical stretch of y x

by a factor of 2.

(e) G; y x is a reflection of

y x across the x-axis.

(f) A; y x is a reflection of y x

across the y-axis.

(g) E; 2y x is a reflection of 2y x

across the x-axis. 2y x is a vertical

stretch of y x by a factor of 2.

(h) I; 2 2y x is a shift of y x 2

units to the right and 2 units upward.

(i) B; 2 2y x is a shift of y x 2

units to the left and 2 units downward.

16. The graph of 32 1 6f x x is the graph

of 3f x x stretched vertically by a factor

of 2, shifted left 1 unit and down 6 units.

17. 3f x x

x h x x 3f x x

−2 2 6

−1 1 3

0 0 0

1 1 3

2 2 6

18. 4f x x

x h x x 4f x x

−2 2 8

−1 1 4

0 0 0

1 1 4

2 2 8

19. 23

f x x

x h x x 23

f x x

−3 3 2

−2 2 43

−1 1 23

0 0 0

1 1 23

2 2 43

3 3 2

20. 34

f x x

x h x x 34

f x x

−4 4 3

−3 3 94

−2 2 32

−1 1 34

0 0 0

1 1 34

2 2 32

3 3 94

4 4 3




(continued)

21. 22f x x

x 2h x x 22f x x

−2 4 8

−1 1 2

0 0 0

1 1 2

2 4 8

22. 23f x x

x 2h x x 23f x x

−2 4 12

−1 1 3

0 0 0

1 1 3

2 4 12

23. 212

f x x

x 2h x x 212

f x x

−2 4 2

−1 1 12

0 0 0

1 1 12

2 4 2

24. 213

f x x

x 2h x x 213

f x x

−3 9 3

−2 4 43

−1 1 13

0 0 0

1 1 13

2 4 43

3 9 3



25. 212

f x x

x 2h x x 212

f x x

−3 9 92

−2 4 2

−1 1 12

0 0 0

1 1 12

2 4 2

3 9 92

26. 213

f x x

x 2h x x 213

f x x

−3 9 −3

−2 4 43

−1 1 13

0 0 0

1 1 13

2 4 43

3 9 −3

27. 3f x x

x h x x 3f x x

−2 2 −6

−1 1 −3

0 0 0

1 1 −3

2 2 −6

28. 2f x x

x h x x 2f x x

−2 2 −4

−1 1 −2

0 0 0

1 1 −2

2 2 −4

29. 12

h x x

x f x x 1

21 12 2

h x x

x x

−4 4 2

−3 3 32

−2 2 1

−1 1 12

0 0 0




(continued)

x f x x 1

21 12 2

h x x

x x

1 1 12

2 2 1

3 3 32

4 4 2

30. 13

h x x

x 13

f x x 1

31 13 3

h x x

x x

−3 3 1

−2 2 23

−1 1 13

0 0 0

1 1 13

2 2 23

3 3 1

31. 4h x x

x f x x 4 2h x x x

0 0 0

1 1 2

2 2 2 2

x f x x 4 2h x x x

3 3 2 3

4 2 4

32. 9h x x

x f x x 9 3h x x x

0 0 0

1 1 3

2 2 3 2

3 3 3 3

4 2 6

33. f x x

x h x x f x x

−4 2 −2

−3 3 3

−2 2 2

−1 1 −1

0 0 0



34. f x x

x h x x f x x

−3 3 −3

−2 2 −2

−1 1 −1

0 0 0

1 1 −1

2 2 −2

3 3 −3

35. (a) 4y f x is a horizontal translation

of f, 4 units to the left. The point that corresponds to (8, 12) on this translated function would be 8 4,12 4,12 .

(b) 4y f x is a vertical translation of f, 4 units up. The point that corresponds to (8, 12) on this translated function would be 8,12 4 8,16 .

36. (a) 14

y f x is a vertical shrinking of f, by

a factor of 14

. The point that corresponds

to (8, 12) on this translated function

would be 14

8, 12 8,3 .

(b) 4y f x is a vertical stretching of f, by

a factor of 4. The point that corresponds to (8, 12) on this translated function would be 8, 4 12 8, 48 .

37. (a) (4 )y f x is a horizontal shrinking of f, by a factor of 4. The point that corresponds to (8, 12) on this translated

function is 14

8 , 12 2, 12 .

(b) 14

y f x is a horizontal stretching of f, by a factor of 4. The point that corresponds to (8, 12) on this translated function is 8 4, 12 32, 12 .

38. (a) The point that corresponds to (8, 12) when reflected across the x-axis would be (8, −12).

(b) The point that corresponds to (8, 12) when reflected across the y-axis would be (−8, 12).

39. (a) The point that is symmetric to (5, –3) with respect to the x-axis is (5, 3).

(b) The point that is symmetric to (5, –3) with respect to the y-axis is (–5, –3).

(c) The point that is symmetric to (5, –3) with respect to the origin is (–5, 3).

40. (a) The point that is symmetric to (–6, 1) with respect to the x-axis is (–6, –1).

(b) The point that is symmetric to (–6, 1) with respect to the y-axis is (6, 1).

(c) The point that is symmetric to (–6, 1) with respect to the origin is (6, –1).

41. (a) The point that is symmetric to (–4, –2) with respect to the x-axis is (–4, 2).

(b) The point that is symmetric to (–4, –2) with respect to the y-axis is (4, –2).

(c) The point that is symmetric to (–4, –2) with respect to the origin is (4, 2).



42. (a) The point that is symmetric to (–8, 0) with respect to the x-axis is (–8, 0) because this point lies on the x-axis.

(b) The point that is symmetric to the point (–8, 0) with respect to the y-axis is (8, 0).

(c) The point that is symmetric to the point (–8, 0) with respect to the origin is (8, 0).

43. The graph of y = |x − 2| is symmetric with respect to the line x = 2.

44. The graph of y = −|x + 1| is symmetric with respect to the line x = −1.

45. 2 5y x Replace x with –x to obtain

2 2( ) 5 5y x x . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain

2 2 2( ) 2 2 2.y x y x y x The result is not the same as the original

equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.


4 42( ) 3 2 3y x x The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to

obtain 4 4– 2( ) 3 2 3y x y x 42 3y x . The result is not the same as

the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.

47. 2 2 12x y Replace x with –x to obtain

2 2 2 2( ) 12 12x y x y . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain

2 2 2 2( ) 12 12x y x y The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin.

48. 2 2 6y x Replace x with –x to obtain

22 2 26 6y x y x

The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain

2 2 2 2( ) 6 6y x y x The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the x-axis, the y-axis, and the origin.

49. 34y x x Replace x with –x to obtain

3 3

3

4( ) ( ) 4( )

4 .

y x x y x xy x x

The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to

obtain 3 34 4 .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain

3 3

3 34( ) ( ) 4( )

4 4 .

y x x y x xy x x y x x

The result is the same as the original equation, so the graph is symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the origin only.




3 3( ) ( ) .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to

obtain 3 3 .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to

obtain 3 3( ) ( )y x x y x x 3 .y x x The result is the same as the

original equation, so the graph is symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the origin only.

51. 2 8y x x Replace x with –x to obtain

2 2( ) ( ) 8 8.y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y

to obtain 2( ) ( ) 8y x x 2 28 8.y x x y x x

The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.

52. y = x + 15 Replace x with –x to obtain

( ) 15 15.y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain ( ) 15 15.y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.

53. 3 2f x x x

3

3 3

2

2 2

f x x xx x x x f x

The function is odd.

54. 5 32f x x x

5 3

5 3 5 3

2

2 2

f x x xx x x x f x


55. 4 20.5 2 6f x x x

4 2

4 2

0.5 2 6

0.5 2 6

f x x xx x f x

The function is even.

56. 20.75 4f x x x

2

2

0.75 4

0.75 4

f x x xx x f x


57. 3 9f x x x

3

3 3

9

9 9

f x x xx x x x f x

The function is neither.

58. 4 5 8f x x x

4

4

5 8

5 8

f x x xx x f x


59. 2 1f x x

This graph may be obtained by translating the

graph of 2y x 1 unit downward.

60. 2 2f x x


graph of 2y x 2 units downward.



61. 2 2f x x


graph of 2y x 2 units upward.

62. 2 3f x x



63. 24g x x


graph of 2y x 4 units to the right.

64. 22g x x



65. 22g x x


graph of 2y x 2 units to the left.

66. 23g x x



67. 1g x x

The graph is obtained by translating the graph of y x 1 unit downward.

68. 3 2g x x

This graph may be obtained by translating the graph of y x 3 units to the left and 2 units

upward.



69. 3( 1)h x x


graph of 3y x 1 unit to the left. It is then reflected across the x-axis.

70. 3( 1)h x x

This graph can be obtained by translating the

graph of 3y x 1 unit to the right. It is then reflected across the x-axis. (We may also reflect the graph about the x-axis first and then translate it 1 unit to the right.)

71. 22 1h x x


graph of 2y x 1 unit down. It is then stretched vertically by a factor of 2.

72. 23 2h x x

This graph may be obtained by stretching the

graph of 2y x vertically by a factor of 3, then shifting the resulting graph down 2 units.

73. 22( 2) 4f x x


graph of 2y x 2 units to the right and 4 units down. It is then stretched vertically by a factor of 2.

74. 23( 2) 1f x x


graph of 2y x 2 units to the right and 1 unit up. It is then stretched vertically by a factor of 3 and reflected over the x-axis.

75. 2f x x


graph of y x two units to the left.



76. 3f x x


graph of y x three units to the right.

77. f x x

This graph may be obtained by reflecting the

graph of y x across the x-axis.

78. 2f x x


graph of y x two units down.

79. 2 1f x x


graph of y x vertically by a factor of two and then translating the resulting graph one unit up.

80. 3 2f x x


graph of y x vertically by a factor of three and then translating the resulting graph two units down.

81. 312

4g x x


graph of 3y x vertically by a factor of 12

,

then shifting the resulting graph down four units.

82. 312

2g x x



,

then shifting the resulting graph up two units.



83. 33g x x

This graph may be obtained by shifting the

graph of 3y x three units left.

84. 32f x x


graph of 3y x two units right.

85. 223

2f x x


graph of 2y x two units to the right, then stretching the resulting graph vertically by a

factor of 23

.

86. Because ( ) ( ),g x x x f x the graphs

are the same.

87. (a) y = g(–x) The graph of g(x) is reflected across the y-axis.

(b) y = g(x – 2) The graph of g(x) is translated to the right 2 units.

(c) y = –g(x) The graph of g(x) is reflected across the x-axis.

(d) y = –g(x) + 2 The graph of g(x) is reflected across the x-axis and translated 2 units up.



88. (a) y f x

The graph of f(x) is reflected across the x-axis.

(b) 2y f x

The graph of f(x) is stretched vertically by a factor of 2.

(c) y f x

The graph of f(x) is reflected across the y-axis.

(d) 12

y f x

The graph of f(x) is compressed vertically

by a factor of 12

.

89. It is the graph of f x x translated 1 unit to

the left, reflected across the x-axis, and translated 3 units up. The equation is

1 3.y x

90. It is the graph of g x x translated 4 units

to the left, reflected across the x-axis, and translated two units up. The equation is

4 2.y x

91. It is the graph of f x x translated one

unit right and then three units down. The

equation is 1 3.y x

92. It is the graph of f x x translated 2 units

to the right, shrunken vertically by a factor of 12

, and translated one unit down. The

equation is 12

2 1.y x


to the left, stretched vertically by a factor of 2, and translated four units down. The equation

is 2 4 4.y x

94. It is the graph of f x x reflected across

the x-axis and then shifted two units down. The equation is 2y x .

95. Because f(3) = 6, the point (3, 6) is on the graph. Because the graph is symmetric with respect to the origin, the point (–3, –6) is on the graph. Therefore, f(–3) = –6.

96. Because f(3) = 6, (3, 6) is a point on the graph. The graph is symmetric with respect to the y-axis, so (–3, 6) is on the graph. Therefore, f(–3) = 6.

97. Because f(3) = 6, the point (3, 6) is on the graph. The graph is symmetric with respect to the line x = 6 and the point (3, 6) is 3 units to the left of the line x = 6, so the image point of (3, 6), 3 units to the right of the line x = 6 is (9, 6). Therefore, f(9) = 6.

98. Because f(3) = 6 and f(–x) = f(x), f(–3) = f(3). Therefore, f(–3) = 6.

99. Because (3, 6) is on the graph, (–3, –6) must also be on the graph. Therefore, f(–3) = –6.

100. If f is an odd function, f(–x) = –f(x). Because f(3) = 6 and f(–x) = –f(x), f(–3) = –f(3). Therefore, f(–3) = –6.

Chapter 2 Quiz (Sections 2.5−2.7) 237


101. f(x) = 2x + 5 Translate the graph of ( )f x up 2 units to obtain the graph of

( ) (2 5) 2 2 7.t x x x Now translate the graph of t(x) = 2x + 7 left 3 units to obtain the graph of

( ) 2( 3) 7 2 6 7 2 13.g x x x x (Note that if the original graph is first translated to the left 3 units and then up 2 units, the final result will be the same.)

102. f(x) = 3 – x Translate the graph of ( )f x down 2 units to

obtain the graph of ( ) (3 ) 2 1.t x x x

Now translate the graph of 1t x x right

3 units to obtain the graph of ( ) ( 3) 1 3 1 4.g x x x x

(Note that if the original graph is first translated to the right 3 units and then down 2 units, the final result will be the same.)

103. (a) Because f(–x) = f(x), the graph is symmetric with respect to the y-axis.

(b) Because f(–x) = –f(x), the graph is symmetric with respect to the origin.

104. (a) f(x) is odd. An odd function has a graph symmetric with respect to the origin. Reflect the left half of the graph in the origin.

(b) f(x) is even. An even function has a graph symmetric with respect to the y-axis. Reflect the left half of the graph in the y-axis.


1. (a) First, find the slope: 9 5

21 ( 3)

m

Choose either point, say, (−3, 5), to find the equation of the line:

5 2( ( 3)) 2( 3) 5y x y x 2 11y x .

(b) To find the x-intercept, let y = 0 and solve

for x: 112

0 2 11x x . The

x-intercept is 112

, 0 .

2. Write 3x − 2y = 6 in slope-intercept form to

find its slope: 32

3 2 6 3.x y y x

Then, the slope of the line perpendicular to

this graph is 23

. 23

4 ( ( 6))y x

2 23 3

( 6)) 4y x y x

3. (a) 8x (b) 5y

4. (a) Cubing function; domain: ( , ) ;

range: ( , ) ; increasing over

( , ) .

(b) Absolute value function; domain: ( , ) ; range: [0, ) ; decreasing over

( , 0) ; increasing over (0, )



(c) Cube root function: domain: ( , ) ;


( , ) .

5.

0.40 0.75

5.5 0.40 5.5 0.75

0.40 5 0.75 2.75

f x xf

A 5.5-minute call costs $2.75.

6. if 0( )2 3 if 0

x xf xx x

For values of x < 0, the graph is the line y = 2x + 3. Do not include the right endpoint

(0, 3). Graph the line y x for values of x ≥ 0, including the left endpoint (0, 0).

7. 3( ) 1f x x

Reflect the graph of 3( )f x x across the x-axis, and then translate the resulting graph one unit up.

8. ( ) 2 1 3f x x

Shift the graph of ( )f x x one unit right,

stretch the resulting graph vertically by a factor of 2, then shift this graph three units up.

9. This is the graph of ( )g x x , translated four units to the left, reflected across the x-axis, and then translated two units down.

The equation is 4 2y x .

10. (a) 2 7f x x

Replace x with –x to obtain

2

2

( ) 7

7

f x xf x x f x

The result is the same as the original function, so the function is even.

(b) 3 1f x x x


3

3

1

1

f x x xx x f x

The result is not the same as the original equation, so the function is not even. Because ,f x f x the function is

not odd. Therefore, the function is neither even nor odd.

(c) 101 99f x x x


101 99

101 99

101 99

f x x xx x

x xf x

Because f(−x) = −f(x), the function is odd.

Section 2.8 Function Operations and Composition

In exercises 1–10, 1f x x and 2.g x x

1. 2

2 2 2

2 1 2 7

f g f g

2. 2

2 2 2

2 1 2 1

f g f g

3. 2

2 2 2

2 1 2 12

fg f g

4. 2

2 2 1 32

2 42

ffg g

5. 2 22 2 2 2 1 5f g f g f

Section 2.8 Function Operations and Composition 239


6. 22 2 2 1 2 1 9g f g f g

7. f is defined for all real numbers, so its domain is , .

8. g is defined for all real numbers, so its domain is , .

9. f + g is defined for all real numbers, so its domain is , .

10. fg

is defined for all real numbers except those

values that make 0,g x so its domain is

, 0 0, .

In Exercises 11–18, 2( ) 3f x x and

( ) 2 6g x x .

11.

2

( )(3) (3) (3)

(3) 3 2(3) 6

12 0 12

f g f g

12. 2

( )( 5) ( 5) ( 5)

[( 5) 3] [ 2( 5) 6]28 16 44

f g f g

13. 2

( )( 1) ( 1) ( 1)

[( 1) 3] [ 2( 1) 6]4 8 4

f g f g

14. 2

( )(4) (4) (4)

[(4) 3] [ 2(4) 6]19 ( 2) 21

f g f g

15. 2

( )(4) (4) (4)

[4 3] [ 2(4) 6]19 ( 2) 38

fg f g

16. 2

( )( 3) ( 3) ( 3)

[( 3) 3] [ 2( 3) 6]12 12 144

fg f g

17. 2( 1) ( 1) 3 4 1

( 1)( 1) 2( 1) 6 8 2

f fg g

18. 2(5) (5) 3 28

(5) 7(5) 2(5) 6 4

f fg g

19. ( ) 3 4, ( ) 2 5f x x g x x

( )( ) ( ) ( )(3 4) (2 5) 5 1

f g x f x g xx x x

( )( ) ( ) ( )(3 4) (2 5) 9

f g x f x g xx x x

2

2

( )( ) ( ) ( ) (3 4)(2 5)

6 15 8 20

6 7 20

fg x f x g x x xx x xx x

( ) 3 4( )

( ) 2 5

f f x xxg g x x

The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g,

and fg are all , . The domain of fg is

the set of all real numbers for which 0.g x This is the set of all real numbers

except 52

, which is written in interval notation

as 5 52 2

, , .

20. f(x) = 6 – 3x, g(x) = –4x + 1 ( )( ) ( ) ( )

(6 3 ) ( 4 1)7 7

f g x f x g xx x

x

( )( ) ( ) ( )(6 3 ) ( 4 1) 5

f g x f x g xx x x

2

2

( )( ) ( ) ( ) (6 3 )( 4 1)

24 6 12 3

12 27 6

fg x f x g x x xx x x

x x

( ) 6 3( )

( ) 4 1

f f x xxg g x x

The domains of both f and g are the set of all real numbers, so the domains of f + g, f – g, and fg are all , . The domain of

fg is the set of all real numbers for which

0.g x This is the set of all real numbers

except 14


as 1 14 4

– , , .

21. 2 2( ) 2 3 , ( ) 3f x x x g x x x

2 2

2

( )( ) ( ) ( )

(2 3 ) ( 3)

3 4 3

f g x f x g xx x x x

x x

2 2

2 2

2

( )( ) ( ) ( )

(2 3 ) ( 3)

2 3 3

2 3


x x x xx x




(continued)

2 2

4 3 2 3 2

4 3 2

( )( ) ( ) ( )

(2 3 )( 3)

2 2 6 3 3 9

2 5 9 9

fg x f x g xx x x x

x x x x x xx x x x

2

2

( ) 2 3( )

( ) 3

f f x x xxg g x x x



0.g x If 2 3 0x x , then by the

quadratic formula 1 112ix . The equation

has no real solutions. There are no real numbers which make the denominator zero.

Thus, the domain of fg is also , .

22. 2 2( ) 4 2 , ( ) 3 2f x x x g x x x

2 2

2

( )( ) ( ) ( )

(4 2 ) ( 3 2)

5 2


x x

2 2

2 2

2

( )( ) ( ) ( )

(4 2 ) ( 3 2)

4 2 3 2

3 5 2


x x x xx x

2 2

4 3 2 3 2

4 3 2

( )( ) ( ) ( )

(4 2 )( 3 2)

4 12 8 2 6 4

4 10 2 4

fg x f x g xx x x x

x x x x x xx x x x

2

2

( ) 4 2( )

( ) 3 2

f f x x xxg g x x x


and fg are all , . The domain of fg is the

set of all real numbers x such that 2 3 2 0x x . Because 2 3 2 ( 1)( 2)x x x x , the numbers

which give this denominator a value of 0 are

x = 1 and x = 2. Therefore, the domain of fg is

the set of all real numbers except 1 and 2, which is written in interval notation as (– , 1) (1, 2) (2, ) .

23. 1

( ) 4 1, ( )f x x g xx

1( )( ) ( ) ( ) 4 1f g x f x g x x

x

1( )( ) ( ) ( ) 4 1f g x f x g x x

x

( )( ) ( ) ( )

1 4 14 1

fg x f x g xxx

x x

1

( ) 4 1( ) 4 1

( ) x

f f x xx x xg g x

Because 14

4 1 0 4 1 ,x x x the

domain of f is 14

, . The domain of g is

, 0 0, . Considering the intersection

of the domains of f and g, the domains of f + g,

f – g, and fg are all 14

, . Because 1 0x

for any value of x, the domain of fg is also

14

, .

24. 1

( ) 5 4, ( )f x x g xx

( )( ) ( ) ( )1 1

5 4 5 4

f g x f x g x

x xx x

( )( ) ( ) ( )1 1

5 4 5 4

f g x f x g x

x xx x

( )( ) ( ) ( )

1 5 45 4

fg x f x g xxx

x x

1

( ) 5 4( ) 5 4

( ) x

f f x xx x xg g x

Because 45

5 4 0 5 4 ,x x x the

domain of f is 45





, . 1 0x for any

value of x, so the domain of fg is also

45

, .



25. 2008 280 and 2008 470, thus

2008 2008 2008280 470 750 (thousand).

M FT M F

26. 2012 390 and 2012 630, thus

2012 2012 2012390 630 1020 (thousand).

M FT M F

27. Looking at the graphs of the functions, the slopes of the line segments for the period 2008−2012 are much steeper than the slopes of the corresponding line segments for the period 2004−2008. Thus, the number of associate’s degrees increased more rapidly during the period 2008−2012.

28. If 2004 2012, ( ) ,k T k r and F(k) = s, then M(k) = r − s.

29. 2000 2000 200019 13 6

T S T S

It represents the dollars in billions spent for general science in 2000.

30. 2010 2010 201029 11 18

T G T G

It represents the dollars in billions spent on space and other technologies in 2010.

31. Spending for space and other technologies spending decreased in the years 1995−2000 and 2010–2015.

32. Total spending increased the most during the years 2005−2010.

33. (a)

2 2 2

4 2 2

f g f g

(b) ( )(1) (1) (1) 1 ( 3) 4f g f g

(c) ( )(0) (0) (0) 0( 4) 0fg f g

(d) (1) 1 1

(1)(1) 3 3

f fg g

34. (a) ( )(0) (0) (0) 0 2 2f g f g

(b) ( )( 1) ( 1) ( 1)2 1 3

f g f g

(c) ( )(1) (1) (1) 2 1 2fg f g

(d) (2) 4

(2) 2(2) 2

f fg g

35. (a) ( )( 1) ( 1) ( 1) 0 3 3f g f g

(b) ( )( 2) ( 2) ( 2)1 4 5

f g f g

(c) ( )(0) (0) (0) 1 2 2fg f g

(d) (2) 3

(2) undefined(2) 0

f fg g

36. (a) ( )(1) (1) (1) 3 1 2f g f g

(b) ( )(0) (0) (0) 2 0 2f g f g

(c) ( )( 1) ( 1) ( 1) 3( 1) 3fg f g

(d) (1) 3

(1) 3(1) 1

f fg g

37. (a) 2 2 2 7 2 5f g f g

(b) ( )(4) (4) (4) 10 5 5f g f g

(c) ( )( 2) ( 2) ( 2) 0 6 0fg f g

(d) (0) 5

(0) undefined(0) 0

f fg g

38. (a) 2 2 2 5 4 9f g f g

(b) ( )(4) (4) (4) 0 0 0f g f g

(c) ( )( 2) ( 2) ( 2) 4 2 8fg f g

(d) (0) 8

(0) 8(0) 1

f fg g



39. x f x g x f g x f g x fg x f xg

−2 0 6 0 6 6 0 6 6 0 6 0 06

0

0 5 0 5 0 5 5 0 5 5 0 0 50

undefined

2 7 −2 7 2 5 7 2 9 7 2 14 72

3.5

4 10 5 10 5 15 10 5 5 10 5 50 105

2


−2 −4 2 4 2 2 4 2 6 4 2 8 42

2

0 8 −1 8 1 7 8 1 9 8 1 8 81

8

2 5 4 5 4 9 5 4 1 5 4 20 54

1.25

4 0 0 0 0 0 0 0 0 0 0 0 00

undefined

41. Answers may vary. Sample answer: Both the

slope formula and the difference quotient represent the ratio of the vertical change to the horizontal change. The slope formula is stated for a line while the difference quotient is stated for a function f.

42. Answers may vary. Sample answer: As h approaches 0, the slope of the secant line PQ approaches the slope of the line tangent of the curve at P.

43. 2f x x

(a) ( ) 2 ( ) 2f x h x h x h

(b) ( ) ( ) (2 ) (2 )2 2

f x h f x x h xx h x h

(c) ( ) ( )

1f x h f x h

h h

44. 1f x x

(a) ( ) 1 ( ) 1f x h x h x h

(b) ( ) ( ) (1 ) (1 )1 1


(c) ( ) ( )

1f x h f x h

h h

45. 6 2f x x

(a) ( ) 6( ) 2 6 6 2f x h x h x h

(b) ( ) ( )(6 6 2) (6 2)6 6 2 6 2 6

f x h f xx h x

x h x h

(c) ( ) ( ) 6

6f x h f x h

h h

46. 4 11f x x

(a) ( ) 4( ) 11 4 4 11f x h x h x h

(b) ( ) ( )(4 4 11) (4 11)4 4 11 4 11 4

f x h f xx h x

x h x h

(c) ( ) ( ) 4

4f x h f x h

h h

47. 2 5f x x

(a) ( ) 2( ) 52 2 5

f x h x hx h

(b) ( ) ( )( 2 2 5) ( 2 5)

2 2 5 2 5 2

f x h f xx h x

x h x h

(c) ( ) ( ) 2

2f x h f x h

h h



48. ( ) 4 2f x x

(a) ( ) 4( ) 24 4 2

f x h x hx h

(b)

( ) ( )4 4 2 4 24 4 2 4 24

f x h f xx h xx h xh

(c) ( ) ( ) 4

4f x h f x h

h h

49. 1

( )f xx

(a) 1

( )f x hx h

(b)

( ) ( )

1 1

f x h f xx x h

x h x x x hh

x x h

(c)

( ) ( )

1

hx x hf x h f x h

h h hx x h

x x h

50. 2

1( )f x

x

(a) 2

1( )f x h

x h

(b)

22

2 2 22

2 2 2 2

2 22 2

( ) ( )

1 1

2 2

f x h f xx x h

xx h x x hx x xh h xh h

x x h x x h

(c)

2

22

22

22

22

22

( ) ( ) 2

2

2

xh hx x hf x h f x xh h

h h hx x hh x h

hx x hx h

x x h

51. 2( )f x x

(a) 2 2 2( ) ( ) 2f x h x h x xh h

(b) 2 2 2

2( ) ( ) 2

2

f x h f x x xh h xxh h

(c) 2( ) ( ) 2

(2 )

2

f x h f x xh hh h

h x hh

x h

52. 2( )f x x

(a)

2

2 2

2 2

( ) ( )

2

2

f x h x hx xh h

x xh h

(b) 2 2 2

2 2 2

2

( ) ( ) 2

2

2

f x h f x x xh h x

x xh h xxh h

(c) 2( ) ( ) 2

(2 )

2

f x h f x xh hh h

h x hh

x h

53. 2( ) 1f x x

(a) 2

2 2

2 2

( ) 1 ( )

1 ( 2 )

1 2

f x h x hx xh h

x xh h

(b) 2 2 2

2 2 2

2

( ) ( )

(1 2 ) (1 )

1 2 1

2

f x h f xx xh h x

x xh h xxh h

(c) 2( ) ( ) 2

( 2 )

2

f x h f x xh hh h

h x hh

x h

54. 2( ) 1 2f x x

(a) 2

2 2

2 2

( ) 1 2( )

1 2( 2 )

1 2 4 2

f x h x hx xh h

x xh h



(b)

2 2 2

2 2 2

2

( ) ( )

1 2 4 2 1 2

1 2 4 2 1 2

4 2

f x h f xx xh h x

x xh h xxh h

(c) 2( ) ( ) 4 2

(4 2 )

4 2

f x h f x xh hh h

h x hh

x h

55. 2( ) 3 1f x x x

(a) 2

2 2

( ) 3 1

2 3 3 1

f x h x h x hx xh h x h

(b)

2 2

2

2 2 2

2

( ) ( )

2 3 3 1

3 1

2 3 3 1 3 1

2 3

f x h f xx xh h x h

x x

x xh h x h x xxh h h

(c) 2( ) ( ) 2 3

(2 3)

2 3

f x h f x xh h hh h

h x hh

x h

56. 2( ) 4 2f x x x

(a) 2

2 2

( ) 4 2

2 4 4 2


(b)

2 2

2

2 2 2

2

( ) ( )

2 4 4 2

4 2

2 4 4 2 4 2

2 4

f x h f xx xh h x h

x x


(c) 2( ) ( ) 2 4

(2 4)

2 4

f x h f x xh h hh h

h x hh

x h

57. 3 4 4 3 1g x x g

4 4 1

2 1 3 2 3 5

f g f g f

58. 3 2 2 3 1g x x g

2 2 1

2 1 3 2 3 1

f g f g f

59. 3 2 2 3 5g x x g

2 2 5

2 5 3 10 3 7

f g f g f

60. 2 3f x x 3 2 3 3 6 3 3f

3 3 3 3 3 0g f g f g

61. 2 3 0 2 0 3 0 3 3f x x f

0 0 3

3 3 3 3 6

g f g f g

62. 2 3 2 2 2 3 7f x x f

2 2 7

7 3 7 3 10

g f g f g

63. 2 3 2 2 2 3 4 3 1f x x f

2 2 1 2 1 3 1f f f f f

64. 3 2 2 3 5g x x g

2 2 5 5 3 2g g g g g

65. ( )(2) [ (2)] (3) 1f g f g f

66. ( )(7) [ (7)] (6) 9f g f g f

67. ( )(3) [ (3)] (1) 9g f g f g

68. ( )(6) [ (6)] (9) 12g f g f g

69. ( )(4) [ (4)] (3) 1f f f f f

70. ( )(1) [ (1)] (9) 12g g g g g

71. ( )(1) [ (1)] (9)f g f g f However, f(9) cannot be determined from the table given.

72. ( ( ))(7) ( ( (7)))( (6)) (9) 12

g f g g f gg f g

73. (a) ( )( ) ( ( )) (5 7)6(5 7) 930 42 9 30 33

f g x f g x f xx

x x

The domain and range of both f and g are ( , ) , so the domain of f g is

( , ) .



(b) ( )( ) ( ( )) ( 6 9)5( 6 9) 7

30 45 7 30 52

g f x g f x g xx

x x

The domain of g f is ( , ) .

74. (a) ( )( ) ( ( )) (3 1)8(3 1) 1224 8 12 24 4

f g x f g x f xxx x


( , ) .

(b) ( )( ) ( ( )) (8 12)3(8 12) 124 36 1 24 35

g f x g f x g xxx x


75. (a) ( )( ) ( ( )) ( 3) 3f g x f g x f x x

The domain and range of g are ( , ) , however, the domain and range of f are [0, ) . So, 3 0 3x x .

Therefore, the domain of f g is

[ 3, ) .

(b) ( )( ) ( ( )) 3g f x g f x g x x

The domain and range of g are ( , ) , however, the domain and range of f are [0, ) .Therefore, the domain of g f is

[0, ) .

76. (a) ( )( ) ( ( )) ( 1) 1f g x f g x f x x

The domain and range of g are ( , ) , however, the domain and range of f are [0, ) . So, 1 0 1x x . Therefore,

the domain of f g is [1, ) .

(b) ( )( ) ( ( )) 1g f x g f x g x x

The domain and range of g are ( , ), however, the domain and range of f are [0, ). Therefore, the domain of g f is

[0, ).

77. (a) 2

2 3

( )( ) ( ( )) ( 3 1)

( 3 1)

f g x f g x f x xx x

The domain and range of f and g are ( , ), so the domain of f g is

( , ).

(b)

3

23 3

6 3

( )( ) ( ( )) ( )

3 1

3 1

g f x g f x g x

x x

x x

The domain and range of f and g are ( , ), so the domain of g f is

( , ).

78. (a) 4 2

4 2

4 2

( )( ) ( ( )) ( 4)

4 2

2

f g x f g x f x xx xx x

The domain of f and g is ( , ), while

the range of f is ( , ) and the range of

g is 4, , so the domain of f g is

( , ).

(b) 4 2

( )( ) ( ( )) ( 2)

( 2) ( 2) 4

g f x g f x g xx x



g is 4, , so the domain of g f is

( , ).

79. (a) ( )( ) ( ( )) (3 ) 3 1f g x f g x f x x

The domain and range of g are ( , ) ,

however, the domain of f is [1, ), while

the range of f is [0, ). So, 13

3 1 0x x . Therefore, the

domain of f g is 13

, .

(b) ( )( ) ( ( )) 1

3 1

g f x g f x g x

x


however, the range of f is [0, ) . So

1 0 1x x . Therefore, the domain of g f is [1, ) .

80. (a) ( )( ) ( ( )) (2 ) 2 2f g x f g x f x x


however, the domain of f is [2, ) . So,

2 2 0 1x x . Therefore, the

domain of f g is 1, .



(b) ( )( ) ( ( )) 2g f x g f x g x

2 2x




81. (a) 21

( )( ) ( ( )) ( 1) xf g x f g x f x

The domain and range of g are ( , ) , however, the domain of f is ( ,0) (0, ) . So, 1 0 1x x .


( , 1) ( 1, ) .

(b) 2 2( )( ) ( ( )) 1x xg f x g f x g

The domain and range of f is ( ,0) (0, ) , however, the domain

and range of g are ( , ) . So 0x .

Therefore, the domain of g f is

( ,0) (0, ) .

82. (a) 44

( )( ) ( ( )) ( 4) xf g x f g x f x



( , 4) ( 4, ) .

(b) 4 4( )( ) ( ( )) 4x xg f x g f x g




( ,0) (0, ) .

83. (a) 1 1( )( ) ( ( )) 2x xf g x f g x f

The domain and range of g are ( , 0) (0, ) , however, the domain

of f is [ 2, ) . So, 1 2 0x

12

0 or x x (using test intervals).


12

, 0 , .

(b) 12

( )( ) ( ( )) 2x

g f x g f x g x

The domain of f is [ 2, ) and its range is

[0, ). The domain and range of g

are ( ,0) (0, ) . So 2 0 2x x .

Therefore, the domain of g f is ( 2, ) .

84. (a) 2 2( )( ) ( ( )) 4x xf g x f g x f


of f is [ 4, ) . So, 2 4 0x

12



12

, 0 , .

(b) 24

( )( ) ( ( )) 4x

g f x g f x g x



are ( , 0) (0, ) . So 4 0 4x x .


85. (a) 1 15 5

( )( ) ( ( )) x xf g x f g x f

The domain of g is ( , 5) ( 5, ) ,

and the range of g is ( , 0) (0, ) .

The domain of f is [0, ) . Therefore, the


(b) 15

( )( ) ( ( ))x

g f x g f x g x

The domain and range of f is [0, ). The

domain of g is ( , 5) ( 5, ).

Therefore, the domain of g f is [0, ).

86. (a) 3 36 6

( )( ) ( ( )) x xf g x f g x f





(b) 36

( )( ) ( ( ))x

g f x g f x g x

The domain and range of f is [0, ) . The

domain of g is ( , 6) ( 6, ) .




87. (a) 1 11 2 1 2

( )( ) ( ( )) xx x xf g x f g x f

The domain and range of g are ( , 0) (0, ) . The domain of f is

( , 2) ( 2, ), and the range of f is

( , 0) (0, ) . So, 1 2

0 0xx x or

12

0 x or 12

x (using test intervals).

Thus, 0x and 12

x . Therefore, the

domain of f g is

1 12 2

, 0 0, , .

(b) 1 12 1 ( 2)

( )( ) ( ( ))

2x xg f x g f x g

x

The domain and range of g are ( , 0) (0, ). The domain of f is

( , 2) (2, ), and the range of f is

( , 0) (0, ). Therefore, the domain of

g f is ( , 2) (2, ).

88. (a) 1 11 4

1 4

( )( ) ( ( )) x xx

x

f g x f g x f



( , 0) (0, ). So, 1 4

0 0xx x

or 14

0 x or 14

1 4 0x x

(using test intervals). Thus, x ≠ 0 and

x ≠ 14

. Therefore, the domain of f g is

1 14 4

, 0 0, , .

(b) 1 14 1 ( 4)

( )( ) ( ( ))

4x xg f x g f x g

x

The domain and range of g are ( ,0) (0, ) . The domain of f is

( , 4) ( 4, ) , and the range of f is

( ,0) (0, ) . Therefore, the domain of

g f is ( , 4) ( 4, ).

89.

(2) (1) 2 and (3) (2) 5

Since (1) 7 and (1) 3, (3) 7.

g f g g f gg f f g

x f x g x ( )g f x

1 3 2 7

2 1 5 2

3 2 7 5

90. ( )f x is odd, so ( 1) (1) ( 2) 2.f f

Because ( )g x is even, (1) ( 1) 2g g and

(2) ( 2) 0.g g ( )( 1) 1,f g so

( 1) 1f g and (2) 1.f ( )f x is odd, so

( 2) (2) 1.f f Thus,

( )( 2) ( 2) (0) 0f g f g f and

( )(1) (1) (2) 1f g f g f and

( )(2) (2) (0) 0.f g f g f

x –2 –1 0 1 2

f x –1 2 0 –2 1

g x 0 2 1 2 0

f g x 0 1 –2 1 0

91. Answers will vary. In general, composition of functions is not commutative. Sample answer:

2 3 3 2 3 26 9 2 6 11

3 2 2 3 2 36 4 3 6 7

f g x f x xx x

g f x g x xx x

Thus, f g x is not equivalent to

g f x .

92.

3

33

7

7 7

7 7

f g x f g x f x

xx x

3

33 33

7

7 7

g f x g f x g x

x x x

93.

14

14

4 2 2

4 2 2

2 2 2 2

f g x f g x x

xx x x

14

1 14 4

4 2 2

4 2 2 4

g f x g f x xx x x

94.

13

13

3

3

f g x f g x x

x x

13

13

3

3

g f x g f x x

x x



95. 31 435 5

3 33 3

5 4

4 4

f g x f g x x

x x x

31 435 5

51 4 4 45 5 5 5 555

5 4

5 4 x

x

g f x g f x x

xx

96. 33

3 33 3

1 1

1 1

f g x f g x x

x x x

33 1 1

1 1

g f x g f x xx x

In Exercises 97−102, we give only one of many possible answers.

97. 2( ) (6 2)h x x

Let g(x) = 6x – 2 and 2( )f x x . 2( )( ) (6 2) (6 2) ( )f g x f x x h x

98. 2 2( ) (11 12 )h x x x 2 2Let ( ) 11 12 and ( ) .g x x x f x x

2

2 2

( )( ) (11 12 )

(11 12 ) ( )

f g x f x xx x h x

99. 2( ) 1h x x

Let 2( ) 1 and ( ) .g x x f x x

2 2( )( ) ( 1) 1 ( ).f g x f x x h x

100. 3( ) (2 3)h x x 3Let ( ) 2 3 and ( ) .g x x f x x

3( )( ) (2 3) (2 3) ( )f g x f x x h x

101. ( ) 6 12h x x

Let ( ) 6 and ( ) 12.g x x f x x

( )( ) (6 ) 6 12 ( )f g x f x x h x

102. 3( ) 2 3 4h x x

Let 3( ) 2 3 and ( ) 4.g x x f x x 3( )( ) (2 3) 2 3 4 ( )f g x f x x h x

103. f(x) = 12x, g(x) = 5280x ( )( ) [ ( )] (5280 )

12(5280 ) 63,360f g x f g x f x

x x

The function f g computes the number of inches in x miles.

104. 3 , 1760f x x g x x

1760

3 1760 5280

f g x f g x f xx x

f g x compute the number of feet in x

miles.

105. 23( )

4x x

(a) 2 2 23 3(2 ) (2 ) (4 ) 3

4 4x x x x

(b) 2(16) (2 8) 3(8)

64 3 square units

A

106. (a) 4 4

4 x xx s s s

(b) 2224 16x xy s

(c) 26 36

2.25 square units16 16

y

107. (a) 2( ) 4 and ( )r t t r r

2 2( )( ) [ ( )]

(4 ) (4 ) 16

r t r tt t t

(b) ( )( )r t defines the area of the leak in terms of the time t, in minutes.

(c) 2 2(3) 16 (3) 144 ft

108. (a) 2 2

( )( ) [ ( )]

(2 ) (2 ) 4

r t r tt t t

(b) It defines the area of the circular layer in terms of the time t, in hours.

(c) 2 2( )(4) 4 (4) 64 mir

109. Let x = the number of people less than 100 people that attend.

(a) x people fewer than 100 attend, so 100 – x people do attend N(x) = 100 – x

(b) The cost per person starts at $20 and increases by $5 for each of the x people that do not attend. The total increase is $5x, and the cost per person increases to $20 + $5x. Thus, G(x) = 20 + 5x.

(c) ( ) ( ) ( ) (100 )(20 5 )C x N x G x x x

Chapter 2 Review Exercises 249


(d) If 80 people attend, x = 100 – 80 = 20.

20 100 20 20 5 20

80 20 100

80 120 $9600

C

110. (a) 1 0.02y x

(b) 2 0.015( 500)y x

(c) 1 2y y represents the total annual interest.

(d) 1 2

1 2

( )(250)(250) (250)

0.02(250) 0.015(250 500)5 0.015(750) 15 11.25$16.25

y yy y

111. (a) 1

2g x x

(b) 1f x x

(c) 1 11

2 2f g x f g x f x x

(d) 160 60 1 $31

2f g

112. If the area of a square is 2x square inches, each side must have a length of x inches. If 3 inches is added to one dimension and 1 inch is subtracted from the other, the new dimensions will be x + 3 and x – 1. Thus, the area of the resulting rectangle is (x) = (x + 3)(x – 1).

Chapter 2 Review Exercises

1. P(3, –1), Q(–4, 5) 2 2

2 2

( , ) ( 4 3) [5 ( 1)]

( 7) 6 49 36 85

d P Q

Midpoint:

3 ( 4) 1 5 1 4 1, , , 2

2 2 2 2 2

2. M(–8, 2), N(3, –7) 2 2

2 2

( , ) [3 ( 8)] ( 7 2)

11 ( 9) 121 81 202

d M N

Midpoint: 8 3 2 ( 7) 5 5

, , 2 2 2 2

3. A(–6, 3), B(–6, 8) 2 2

2

( , ) [ 6 ( 6)] (8 3)

0 5 25 5

d A B

Midpoint: 6 ( 6) 3 8 12 11 11

, , 6,2 2 2 2 2

4. Label the points A(5, 7), B(3, 9), and C(6, 8).

2 2

2 2

( , ) 3 5 9 7

2 2 4 4 8

d A B

2 2

2 2

( , ) (6 5) (8 7)

1 1 1 1 2

d A C

2 2

22

( , ) 6 3 8 9

3 1 9 1 10

d B C


ABC is a right triangle with right angle at (5, 7).

5. Let B have coordinates (x, y). Using the midpoint formula, we have

6 108, 2 ,

2 26 10

8 22 2

6 16 10 422 6

x y

x y

x yx y

The coordinates of B are (22, −6).

6. P(–2, –5), Q(1, 7), R(3, 15) 2 2

2 2

( , ) (1 ( 2)) (7 ( 5))

(3) (12) 9 144

153 3 17

d P Q

2 2

2 2

( , ) (3 1) (15 7)

2 8 4 64 68 2 17

d Q R

2 2

2 2

( , ) (3 ( 2)) (15 ( 5))

(5) (20) 25 400 5 17

d P R




(continued)

( , ) ( , ) 3 17 2 17d P Q d Q R

5 17 ( , ),d P R so these three points are collinear.

7. Center (–2, 3), radius 15 2 2 2

2 2 2

2 2

( ) ( )

[ ( 2)] ( 3) 15

( 2) ( 3) 225

x h y k rx y

x y

8. Center ( 5, 7 ), radius 3

2 2 2

22 2

2 2

( ) ( )

5 7 3

5 7 3

x h y k r

x y

x y

9. Center (–8, 1), passing through (0, 16) The radius is the distance from the center to any point on the circle. The distance between (–8, 1) and (0, 16) is

2 2 2 2(0 ( 8)) (16 1) 8 15

64 225 289 17.

r


2 2[ ( 8)] ( 1) 17

( 8) ( 1) 289

x yx y

10. Center (3, –6), tangent to the x-axis The point (3, –6) is 6 units directly below the x-axis. Any segment joining a circle’s center to a point on the circle must be a radius, so in this case the length of the radius is 6 units.

2 2 2

2 2 2

2 2

( ) ( )

( 3) [ ( 6)] 6

( 3) ( 6) 36

x h y k rx y

x y

11. The center of the circle is (0, 0). Use the distance formula to find the radius:

2 2 2(3 0) (5 0) 9 25 34r

The equation is 2 2 34x y .


2 2 2( 2 0) (3 0) 4 9 13r



2 2 2( 2 0) (6 3) 4 9 13r

The equation is 2 2( 3) 13x y .


2 2 2(4 5) (9 6) 1 9 10r

The equation is 2 2( 5) ( 6) 10x y .

15. 2 24 6 12 0x x y y Complete the square on x and y to put the equation in center-radius form.

2 2

2 2

2 2

4 6 12

4 4 6 9 12 4 9

2 3 1

x x y y

x x y y

x y

The circle has center (2, –3) and radius 1.


2 2

2 2

( 6 9) ( 10 25) 30 9 25

( 3) ( 5) 4

x x y yx y

The circle has center (3, 5) and radius 2.

17.

2 2

2 2

2 2

2 249 9 49 94 4 4 4

2 27 3 49 942 2 4 4 4

2 27 3 542 2 4

2 14 2 6 2 0

7 3 1 0

7 3 1

7 3 1

x x y yx x y yx x y y

x x y y

x y

x y

The circle has center 7 32 2

, and radius

54 9 6 3 6544 24 4

.

18.

2 2

2 2

2 2

2 2 25 25121 1214 4 4 4

22 5 146112 2 4

3 33 3 15 0

11 5 0

11 5 0

11 5 0

x x y yx x y y

x x y y

x x y y

x y

The circle has center 5112 2

, and radius

1462

.

19. This is not the graph of a function because a vertical line can intersect it in two points. domain: [−6, 6]; range: [−6, 6]

20. This is not the graph of a function because a vertical line can intersect it in two points. domain: , ; range: 0,



21. This is not the graph of a function because a vertical line can intersect it in two points. domain: , ; range: ( , 1] [1, )

22. This is the graph of a function. No vertical line will intersect the graph in more than one point. domain: , ; range: 0,

23. This is not the graph of a function because a vertical line can intersect it in two points. domain: 0, ; range: ,

24. This is the graph of a function. No vertical line will intersect the graph in more than one point. domain: , ; range: ,

25. 26y x Each value of x corresponds to exactly one value of y, so this equation defines a function.

26. The equation 213

x y does not define y as a

function of x. For some values of x, there will be more than one value of y. For example, ordered pairs (3, 3) and (3, –3) satisfy the relation. Thus, the relation would not be a function.

27. The equation 2y x does not define y as a function of x. For some values of x, there will be more than one value of y. For example, ordered pairs (3, 1) and (3, −1) satisfy the relation.

28. The equation 4yx

defines y as a function

of x because for every x in the domain, which is (– , 0) (0, ) , there will be exactly one value of y.

29. In the function ( ) 4f x x , we may use

any real number for x. The domain is , .

30. 8

( )8

xf xx

x can be any real number except 8 because this will give a denominator of zero. Thus, the domain is (– , 8) (8, ).

31. 6 3f x x

In the function 6 3y x , we must have

6 3 0x . 6 3 0 6 3 2 2x x x x Thus, the domain is , 2 .

32. (a) As x is getting larger on the interval 2, , the value of y is increasing.

(b) As x is getting larger on the interval , 2 , the value of y is decreasing.

(c) f(x) is constant on (−2, 2).

In exercises 33–36, 22 3 6.f x x x

33. 23 2 3 3 3 62 9 3 3 618 9 6 15

f

34.

20.5 2 0.5 3 0.5 6

2 0.25 3 0.5 60.5 1.5 6 8

f

35. 20 2 0 3 0 6 6f

36. 22 3 6f k k k

37. 25

2 5 5 5 2 5 1x y y x y x

The graph is the line with slope 25

and

y-intercept (0, –)1. It may also be graphed using intercepts. To do this, locate the x-intercept: 0y

52

2 5 0 5 2 5x x x

38. 37

3 7 14 7 3 14 2x y y x y x

The graph is the line with slope of 37

and

y-intercept (0, 2). It may also be graphed using intercepts. To do this, locate the x-intercept by setting y = 0:

143

3 7 0 14 3 14x x x



39. 25

2 5 20 5 2 20 4x y y x y x


and

y-intercept (0, 4). It may also be graphed using intercepts. To do this, locate the x-intercept: x-intercept: 0y

2 5 0 20 2 20 10x x x

40. 13

3y x y x


and

y-intercept (0, 0), which means that it passes through the origin. Use another point such as (6, 2) to complete the graph.

41. f(x) = x The graph is the line with slope 1 and y-intercept (0, 0), which means that it passes through the origin. Use another point such as (1, 1) to complete the graph.

42.

14

4 84 8

2

x yy xy x


and

y-intercept (0, –2). It may also be graphed using intercepts. To do this, locate the x-intercept:

0 4 0 8 8y x x

43. x = –5 The graph is the vertical line through (–5, 0).

44. f(x) = 3 The graph is the horizontal line through (0, 3).

45. 2 0 2y y The graph is the horizontal line through (0, −2).

46. The equation of the line that lies along the x-axis is y = 0.

47. Line through (0, 5), 23

m

Note that 2 23 3

.m

Begin by locating the point (0, 5). Because the

slope is 23 , a change of 3 units horizontally

(3 units to the right) produces a change of –2 units vertically (2 units down). This gives a second point, (3, 3), which can be used to complete the graph.




(continued)

48. Line through (2, –4), 34

m

First locate the point (2, –4).

Because the slope is 34

, a change of 4 units

horizontally (4 units to the right) produces a change of 3 units vertically (3 units up). This gives a second point, (6, –1), which can be used to complete the graph.

49. through (2, –2) and (3, –4) 2 1

2 1

4 2 22

3 2 1

y ymx x

50. through (8, 7) and 12

, 2

2 11 15

2 1 2 2

2 7 9

8

2 18 69

15 15 5

y ymx x

51. through (0, –7) and (3, –7) 7 7 0

03 0 3

m

52. through (5, 6) and (5, –2)

2 1

2 1

2 6 8

5 5 0

y ymx x

The slope is undefined.

53. 11 2 3x y Solve for y to put the equation in slope-intercept form.

3112 2

2 11 3y x y x

Thus, the slope is 112

.

54. 9x – 4y = 2. Solve for y to put the equation in slope-intercept form.

9 14 2

4 9 2y x y x


.

55. 2 0 2x x The graph is a vertical line, through (2, 0). The slope is undefined.

56. x – 5y = 0. Solve for y to put the equation in slope-intercept form.

15

5y x y x


.

57. Initially, the car is at home. After traveling for 30 mph for 1 hr, the car is 30 mi away from home. During the second hour the car travels 20 mph until it is 50 mi away. During the third hour the car travels toward home at 30 mph until it is 20 mi away. During the fourth hour the car travels away from home at 40 mph until it is 60 mi away from home. During the last hour, the car travels 60 mi at 60 mph until it arrived home.

58. (a) This is the graph of a function because no vertical line intersects the graph in more than one point.

(b) The lowest point on the graph occurs in December, so the most jobs lost occurred in December. The highest point on the graph occurs in January, so the most jobs gained occurred in January.

(c) The number of jobs lost in December is approximately 6000. The number of jobs gained in January is approximately 2000.

(d) It shows a slight downward trend.

59. (a) We need to first find the slope of a line that passes between points (0, 30.7) and (12, 82.9)

2 1

2 1

82.9 30.7 52.24.35

12 0 12

y ymx x

Now use the point-intercept form with b = 30.7 and m = 4.35. y = 4.35x + 30.7 The slope, 4.35, indicates that the number of e-filing taxpayers increased by 4.35% each year from 2001 to 2013.

(b) For 2009, we evaluate the function for x = 8. y = 4.35(8) + 30.7 = 65.5 65.5% of the tax returns are predicted to have been filed electronically.



60. We need to find the slope of a line that passes between points (1980, 21000) and (2013, 63800)

2 1

2 1

63,800 21,000

2013 198042,800

$1297 per year33

y ymx x

The average rate of change was about $1297 per year.

61. (a) through (3, –5) with slope –2 Use the point-slope form.

1 1( )( 5) 2( 3)

5 2( 3)5 2 6

2 1

y y m x xy x

y xy x

y x

(b) Standard form: 2 1 2 1y x x y

62. (a) through (–2, 4) and (1, 3) First find the slope.

3 4 1

1 ( 2) 3m

Now use the point-slope form with 1

1 1 3( , ) (1, 3) and .x y m

1 113

1013 3

( )

3 ( 1)3( 3) 1( 1)

3 9 13 10

y y m x xy x

y xy x

y x y x

(b) Standard form: 101

3 33 10

3 10y x y xx y

63. (a) through (2, –1) parallel to 3x – y = 1 Find the slope of 3x – y = 1. 3 1 3 1 3 1x y y x y x The slope of this line is 3. Because parallel lines have the same slope, 3 is also the slope of the line whose equation is to be found. Now use the point-slope form with 1 1( , ) (2, 1) and 3.x y m

1 1( )( 1) 3( 2)

1 3 6 3 7

y y m x xy x

y x y x

(b) Standard form: 3 7 3 7 3 7y x x y x y

64. (a) x-intercept (–3, 0), y-intercept (0, 5) Two points of the line are (–3, 0) and (0, 5). First, find the slope.

5 0 5

0 3 3m

The slope is 53

and the y-intercept is

(0, 5). Write the equation in slope-

intercept form: 53

5y x


5 3 5 155 3 15 5 3 15

y x y xx y x y

65. (a) through (2, –10), perpendicular to a line with an undefined slope A line with an undefined slope is a vertical line. Any line perpendicular to a vertical line is a horizontal line, with an equation of the form y = b. The line passes through (2, –10), so the equation of the line is y = –10.

(b) Standard form: y = –10

66. (a) through (0, 5), perpendicular to 8x + 5y = 3 Find the slope of 8x + 5y = 3.

8 35 5

8 5 3 5 8 3x y y xy x

The slope of this line is 85

. The slope

of any line perpendicular to this line is 58

, because 8 55 8

1.

The equation in slope-intercept form with

slope 58

and y-intercept (0, 5) is

58

5.y x


5 8 5 405 8 40 5 8 40

y x y xx y x y

67. (a) through (–7, 4), perpendicular to y = 8 The line y = 8 is a horizontal line, so any line perpendicular to it will be a vertical line. Because x has the same value at all points on the line, the equation is x = –7. It is not possible to write this in slope-intercept form.

(b) Standard form: x = –7

68. (a) through (3, –5), parallel to y = 4 This will be a horizontal line through (3, –5). Because y has the same value at all points on the line, the equation is y = –5.




69. ( ) 3f x x

The graph is the same as that of ,y x

except that it is translated 3 units downward.

70. ( )f x x

The graph of ( )f x x is the reflection of

the graph of y x about the x-axis.

71. 2( ) 1 3f x x

The graph of 2( ) 1 3f x x is a

translation of the graph of 2y x to the left 1 unit, reflected over the x-axis and translated up 3 units.

72. ( ) 2f x x

The graph of ( ) 2f x x is the reflection

of the graph of y x about the x-axis, translated down 2 units.

73. ( ) 3f x x

To get y = 0, we need 0 3 1x 3 4.x To get y = 1, we need1 3 2x 4 5.x Follow this pattern to graph the step function.

74. 3( ) 2 1 2f x x

The graph of 3( ) 2 1 2f x x is a

translation of the graph of 3y x to the left 1 unit, stretched vertically by a factor of 2, and translated down 2 units.

75. 4 2 if 1( )

3 5 if 1x xf x x x

Draw the graph of y = –4x + 2 to the left of x = 1, including the endpoint at x = 1. Draw the graph of y = 3x – 5 to the right of x = 1, but do not include the endpoint at x = 1. Observe that the endpoints of the two pieces coincide.



76. 2 3 if 2( )

4 if 2x xf x

x x

Graph the curve 2 3y x to the left of x = 2, and graph the line y = –x + 4 to the right of x = 2. The graph has an open circle at (2, 7) and a closed circle at (2, 2).

77. if 3

( )6 if 3x xf x

x x

Draw the graph of y x to the left of x = 3,

but do not include the endpoint. Draw the graph of y = 6 – x to the right of x = 3, including the endpoint. Observe that the endpoints of the two pieces coincide.

78. Because x represents an integer, .x x

Therefore, 2 .x x x x x

79. True. The graph of an even function is symmetric with respect to the y-axis.

80. True. The graph of a nonzero function cannot be symmetric with respect to the x-axis. Such a graph would fail the vertical line test

81. False. For example, 2( )f x x is even and (2, 4) is on the graph but (2, –4) is not.

82. True. The graph of an odd function is symmetric with respect to the origin.

83. True. The constant function 0f x is both

even and odd. Because 0 ,f x f x

the function is even. Also 0 0 ,f x f x so the function is

odd.

84. False. For example, 3( )f x x is odd, and (2, 8) is on the graph but (–2, 8) is not.

85. 2 10x y

Replace x with –x to obtain 2( ) 10.x y The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to

obtain 2 2( ) 10 10.x y x y The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Replace x with –x and y with –y to

obtain 2 2( ) ( ) 10 ( ) 10.x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. The graph is symmetric with respect to the x-axis only.

86. 2 25 5 30y x Replace x with –x to obtain

2 2 2 25 5( ) 30 5 5 30.y x y x The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain

2 2 2 25( ) 5 30 5 5 30.y x y x The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. The graph is symmetric with respect to the y-axis and x-axis, so it must also be symmetric with respect to the origin. Note that

this equation is the same as 2 2 6y x , which is a circle centered at the origin.

87. 2 3x y Replace x with –x to obtain

2 3 2 3( ) .x y x y The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace

y with –y to obtain 2 3 2 3( ) .x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y

with –y to obtain 2 3 2 3( ) ( ) .x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.



88. 3 4y x

Replace x with –x to obtain 3 4y x . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain

3 3( ) 4 4y x y x 3 4y x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain

3 3 3( ) ( ) 4 4 4.y x y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.

89. 6 4x y

Replace x with –x to obtain 6( ) 4x y

6 4x y . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain 6 ( ) 4 6 4x y x y . The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain 6( ) ( ) 4x y 6 4x y . This equation is not equivalent to the original one, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.

90. y x

Replace x with –x to obtain ( ) .y x y x The result is not the

same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain

.y x y x The result is the same as

the original equation, so the graph is symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain

( ) .y x y x The result is not the

same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the x-axis only.

91. y = 1 This is the graph of a horizontal line through (0, 1). It is symmetric with respect to the y-axis, but not symmetric with respect to the x-axis and the origin.

92. x y

Replace x with –x to obtain .x y x y


.x y x y The result is the same as

the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and with respect to the y-axis, it must also by symmetric with respect to the origin.

93. 2 2 0x y Replace x with −x to obtain

2 2 2 20 0.x y x y The result is

the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with −y to obtain

22 2 20 0.x y x y The result is

the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and with respect to the y-axis, it must also by symmetric with respect to the origin.

94. 22 2 4x y

Replace x with −x to obtain

2 2 222 4 2 4.x y x y

The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with −y to obtain

22 2 4.x y The result is not the same

as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain

2 2 222 4 2 4,x y x y

which is not equivalent to the original equation. Therefore, the graph is not symmetric with respect to the origin.

95. To obtain the graph of ( )g x x , reflect the

graph of ( )f x x across the x-axis.

96. To obtain the graph of ( ) 2h x x , translate

the graph of ( )f x x down 2 units.



97. To obtain the graph of ( ) 2 4k x x ,

translate the graph of ( )f x x to the right 4

units and stretch vertically by a factor of 2.

98. If the graph of ( ) 3 4f x x is reflected about the x-axis, we obtain a graph whose equation is (3 4) 3 4.y x x

99. If the graph of ( ) 3 4f x x is reflected about the y-axis, we obtain a graph whose equation is ( ) 3( ) 4 3 4.y f x x x

100. If the graph of ( ) 3 4f x x is reflected about the origin, every point (x, y) will be replaced by the point (–x, –y). The equation for the graph will change from 3 4y x to

3( ) 4y x 3 4y x

3 4.y x

101. (a) To graph ( ) 3,y f x translate the graph of y = f(x), 3 units up.

(b) To graph ( 2),y f x translate the graph of y = f(x), 2 units to the right.

(c) To graph ( 3) 2,y f x translate the graph of y = f(x), 3 units to the left and 2 units down.

(d) To graph ( )y f x , keep the graph of

y = f(x) as it is where 0y and reflect the graph about the x-axis where y < 0.

102. No. For example suppose 2f x x and

2 .g x x Then

( )( ) ( ( )) (2 ) 2 2f g x f g x f x x



2 2 0 1x x . Therefore, the domain of

f g is 1, . The domain of g, ( , ), is

not a subset of the domain of , 1, .f g

For Exercises 103–110, 2( ) 3 4f x x= − and 2( ) 3 4.g x x x= − −

103. 2 2

4 3 2 2

4 3 2

( )( ) ( ) ( )

(3 4)( 3 4)

3 9 12 4 12 16

3 9 16 12 16

fg x f x g xx x x

x x x x xx x x x

104. 2 2

( )(4) (4) (4)

(3 4 4) (4 3 4 4)(3 16 4) (16 3 4 4)(48 4) (16 12 4)44 0 44

f g f g

105. 2 2

( )( 4) ( 4) ( 4)

[3( 4) 4] [( 4) 3( 4) 4][3(16) 4] [16 3( 4) 4][48 4] [16 12 4]44 24 68

f g f g

106. 2 2

2 2

2 2

2

( )(2 ) (2 ) (2 )

[3(2 ) 4] [(2 ) 3(2 ) 4]

[3(4) 4] [4 3(2 ) 4]

(12 4) (4 6 4)

16 6 8

f g k f k g kk k kk k k

k k kk k



107. 2

2

(3) 3 3 4 3 9 4(3)

(3) 9 3 3 43 3 3 427 4 23 23

9 9 4 4 4

f fg g

108.

2

2

3 1 4 3 1 4( 1)

1 3 1 41 3 1 43 4 1

undefined1 3 4 0

fg

109. The domain of (fg)(x) is the intersection of the domain of f(x) and the domain of g(x). Both have domain , , so the domain of

(fg)(x) is , .

110. 2 2

2

3 4 3 4( )

( 1)( 4)3 4

f x xxg x xx x

Because both f x and g x have domain

, , we are concerned about values of x

that make 0.g x Thus, the expression is

undefined if (x + 1)(x – 4) = 0, that is, if x = –1 or x = 4. Thus, the domain is the set of all real numbers except x = –1 and x = 4, or (– , –1) (–1, 4) (4, ).

111. 2 9f x x

( ) 2( ) 9 2 2 9f x h x h x h

( ) ( ) (2 2 9) (2 9)2 2 9 2 9 2


Thus, ( ) ( ) 2

2.f x h f x h

h h

112. 2( ) 5 3f x x x 2

2 2( ) ( ) 5( ) 3

2 5 5 3


2 2 2

2 2 2

2

( ) ( )

( 2 5 5 3) ( 5 3)

2 5 5 3 5 3

2 5

f x h f xx xh h x h x x


2( ) ( ) 2 5

(2 5)2 5

f x h f x xh h hh h

h x h x hh

For Exercises 113–118, 2( ) 2 and ( ) .f x x g x x

113. 2

( )( ) [ ( )] 2

2 2

g f x g f x g x

x x

114. 2 2( )( ) [ ( )] ( ) 2f g x f g x f x x

115. 2, so 3 3 2 1 1.f x x f

Therefore,

3 3 1g f g f g 21 1.

116. 22 , so 6 6 36.g x x g

Therefore, 6 6 36f g f g f

36 2 34 .

117. 1 1 1 2 3g f g f g g

Because 3 is not a real number, 1g f

is not defined.

118. To find the domain of ,f g we must consider the domain of g as well as the composed function, .f g Because

2 2f g x f g x x we need to

determine when 2 2 0.x Step 1: Find the values of x that satisfy

2 2 0.x 2 2 2x x

Step 2: The two numbers divide a number line into three regions.

Step 3 Choose a test value to see if it satisfies

the inequality, 2 2 0.x

Interval Test

Value Is 2 2 0x true or false?

, 2 2 22 2 0 ?2 0 True

2, 2 0 20 2 0 ?

2 0 False

2, 2 22 2 0 ?

2 0 True

The domain of f g is

, 2 2, .



119. 1 1 1 7 1 8f g f g

120. ( )(3) (3) (3) 9 9 0f g f g

121. ( )( 1) ( 1) ( 1) 3 2 6fg f g

122. (0) 5

(0) undefined(0) 0

f fg g

123. ( )( 2) [ ( 2)] (1) 2g f g f g

124. ( )(3) [ (3)] ( 2) 1f g f g f

125. ( )(2) [ (2)] (2) 1f g f g f

126. ( )(3) [ (3)] (4) 8g f g f g

127. Let x = number of yards. f(x) = 36x, where f(x) is the number of inches. g(x) = 1760x, where g(x) is the number of yards. Then

( )( ) ( ) 1760(36 )g f x g f x x 63,360 .x

There are 63,360x inches in x miles

128. Use the definition for the perimeter of a rectangle. P = length + width + length + width

( ) 2 2 6P x x x x x x This is a linear function.

129. If 343

( )V r r and if the radius is increased

by 3 inches, then the amount of volume gained is given by

3 34 43 3

( ) ( 3) ( ) ( 3) .gV r V r V r r r

130. (a) 2V r h If d is the diameter of its top, then h = d

and 2

.dr So,

32 2

2 4 4( ) ( ) ( ) .d d dV d d d

(b) 22 2S r rh

2

2 2 2

2 22 2 2

2 32 2 2

( ) 2 2 ( )d d d

d d d

S d d d

Chapter 2 Test

1. (a) The domain of 3f x x occurs

when 0.x In interval notation, this

correlates to the interval in D, 0, .

(b) The range of 3f x x is all real

numbers greater than or equal to 0. In interval notation, this correlates to the interval in D, 0, .

(c) The domain of 2 3f x x is all real

numbers. In interval notation, this correlates to the interval in C, , .

(d) The range of 2 3f x x is all real

numbers greater than or equal to 3. In interval notation, this correlates to the interval in B, 3, .

(e) The domain of 3 3f x x is all real


(f) The range of 3 3f x x is all real


(g) The domain of 3f x x is all real


(h) The range of 3f x x is all real


(i) The domain of 2x y is 0x because when you square any value of y, the outcome will be nonnegative. In interval notation, this correlates to the interval in D, 0, .

(j) The range of 2x y is all real numbers. In interval notation, this correlates to the interval in C, , .

2. Consider the points 2,1 and 3, 4 .

4 1 3

3 ( 2) 5m

3. We label the points A 2,1 and B 3, 4 .

2 2

2 2

( , ) [3 ( 2)] (4 1)

5 3 25 9 34

d A B

Chapter 2 Test 261


4. The midpoint has coordinates

2 3 1 4 1 5, , .

2 2 2 2

5. Use the point-slope form with 3

1 1 5( , ) ( 2,1) and .x y m

1 13535

( )

1 [ ( 2)]

1 ( 2) 5 1 3( 2)5 5 3 6 5 3 11

3 5 11 3 5 11

y y m x xy xy x y xy x y x

x y x y

6. Solve 3x – 5y = –11 for y.

3 115 5

3 5 115 3 11

x yy xy x

Therefore, the linear function is 3 115 5

( ) .f x x

7. (a) The center is at (0, 0) and the radius is 2, so the equation of the circle is

2 2 4x y .

(b) The center is at (1, 4) and the radius is 1, so the equation of the circle is

2 2( 1) ( 4) 1x y

8. 2 2 4 10 13 0x y x y Complete the square on x and y to write the equation in standard form:

2 2

2 2

2 2

2 2

4 10 13 0

4 10 13

4 4 10 25 13 4 25

2 5 16

x y x yx x y y

x x y y

x y

The circle has center (−2, 5) and radius 4.

9. (a) This is not the graph of a function because some vertical lines intersect it in more than one point. The domain of the relation is [0, 4]. The range is [–4, 4].

(b) This is the graph of a function because no vertical line intersects the graph in more than one point. The domain of the function is (– , –1) (–1, ). The

range is (– , 0) (0, ). As x is getting

larger on the intervals , 1 and

1, , the value of y is decreasing, so

the function is decreasing on these intervals. (The function is never increasing or constant.)

10. Point A has coordinates (5, –3). (a) The equation of a vertical line through A

is x = 5.

(b) The equation of a horizontal line through A is y = –3.

11. The slope of the graph of 3 2y x is –3.

(a) A line parallel to the graph of 3 2y x has a slope of –3.

Use the point-slope form with

1 1( , ) (2,3) and 3.x y m

1 1( )3 3( 2)3 3 6 3 9

y y m x xy xy x y x

(b) A line perpendicular to the graph of

3 2y x has a slope of 13

because

13

3 1.

13

713 3

3 ( 2)

3 3 2 3 9 2

3 7

y xy x y x

y x y x

12. (a) 2, (b) 0, 2

(c) , 0 (d) ,

(e) , (f) 1,

13. To graph 2 1f x x , we translate the

graph of ,y x 2 units to the right and 1 unit

down.



14. ( ) 1f x x

To get y = 0, we need 0 1 1x 1 0.x To get y = 1, we need

1 1 2x 0 1.x Follow this pattern to graph the step function.

15. 12

3 if 2( )

2 if 2x

f x x x

For values of x with x < –2, we graph the horizontal line y = 3. For values of x with

2,x we graph the line with a slope of 12

and a y-intercept of (0, 2). Two points on this line are (–2, 3) and (0, 2).

16. (a) Shift f(x), 2 units vertically upward.

(b) Shift f(x), 2 units horizontally to the left.

(c) Reflect f(x), across the x-axis.

(d) Reflect f(x), across the y-axis.

(e) Stretch f(x), vertically by a factor of 2.

17. Starting with ,y x we shift it to the left 2 units and stretch it vertically by a factor of 2. The graph is then reflected over the x-axis and then shifted down 3 units.

18. 2 23 2 3x y

(a) Replace y with –y to obtain 2 2 2 23 2( ) 3 3 2 3.x y x y

The result is the same as the original equation, so the graph is symmetric with respect to the x-axis.

(b) Replace x with –x to obtain 2 2 2 23( ) 2 3 3 2 3.x y x y

The result is the same as the original equation, so the graph is symmetric with respect to the y-axis.

(c) The graph is symmetric with respect to the x-axis and with respect to the y-axis, so it must also be symmetric with respect to the origin.

Chapter 2 Test 263


19. 2( ) 2 3 2, ( ) 2 1f x x x g x x

(a)

2

2

2

( )( ) ( ) ( )

2 3 2 2 1

2 3 2 2 1

2 1

f g x f x g xx x x

x x xx x

(b) 2( ) 2 3 2

( )( ) 2 1

f f x x xxg g x x

(c) We must determine which values solve the equation 2 1 0.x

12

2 1 0 2 1x x x

Thus, 12

is excluded from the domain,

and the domain is 1 12 2

, , .

(d) 2( ) 2 3 2f x x x

2

2 2

2 2

2 3 2

2 2 3 3 2

2 4 2 3 3 2


x xh h x h

2 2

2

2 2

2

2

( ) ( )

(2 4 2 3 3 2)

(2 3 2)

2 4 2 3

3 2 2 3 2

4 2 3

f x h f xx xh h x h

x xx xh h x

h x xxh h h

2( ) ( ) 4 2 3

(4 2 3)

4 2 3

f x h f x xh h hh h

h x hh

x h

(e) 2

( )(1) (1) (1)

(2 1 3 1 2) ( 2 1 1)(2 1 3 1 2) ( 2 1 1)(2 3 2) ( 2 1)1 ( 1) 0

f g f g

(f) 2

( )(2) (2) (2)

(2 2 3 2 2) ( 2 2 1)(2 4 3 2 2) ( 2 2 1)(8 6 2) ( 4 1)4( 3) 12

fg f g

(g) 2 1 0 2 0 1g x x g

0 1 1. Therefore,

2

0 0

1 2 1 3 1 22 1 3 1 22 3 2 1

f g f gf

For exercises 20 and 21, 1f x x and

2 7.g x x

20. 2 7

(2 7) 1 2 6

f g f g x f xx x

The domain and range of g are ( , ) , while

the domain of f is [0, ) . We need to find the values of x which fit the domain of f: 2 6 0 3x x . So, the domain of f g

is [3, ) .

21. 1

2 1 7

g f g f x g x

x


the domain of f is [0, ) . We need to find the values of x which fit the domain of f:

1 0 1x x . So, the domain of g f

is [ 1, ) .

22. (a) C(x) = 3300 + 4.50x

(b) R(x) = 10.50x

(c) ( ) ( ) ( )10.50 (3300 4.50 )6.00 3300

P x R x C xx x

x

(d) ( ) 06.00 3300 0

6.00 3300550

P xx

xx

She must produce and sell 551 items before she earns a profit.

20 Copyright © 2017 Pearson Education, Inc

Chapter 2 Graphs and Functions Section 2.1 Rectangular Coordinates

and Graphs Classroom Example 1 (page 184)

(a) (transportation, $12,153)

(b) (health care, $4917)

Classroom Example 2 (page 186)

22( , ) 2 3 8 5

25 169 194

d P Q


22

22

2 2

( , ) 5 0 1 2

25 9 34

( , ) 4 0 3 2

16 25 41

( , ) 4 5 3 1

81 4 85

d R S

d R T

d S T

The longest side has length 85

2 2 2?34 41 85

34 41 85

The triangle formed by the three points is not a right triangle.


2 2

2 2

22

The distance between ( 2,5) and (0,3) is

2 0 5 3 4 4 8 2 2

The distance between (0,3) and (8, 5) is

8 0 5 3 64 64

128 8 2

The distance between ( 2,5) and (8, 5) is

2 8 5 5 100 100

200 10 2

P Q

Q R

P R

Because 2 2 8 2 10 2 , the points are collinear.


(a) The coordinates of M are

7 ( 2) 5 13 9

, , 42 2 2

(b) Let (x, y) be the coordinates of Q. Use the midpoint formula to find the coordinates:

8 20, 4, 4

2 2

84 8 8 0

220

4 20 8 122

x y

xx x

yy y

The coordinates of Q are (0, 12).


The year 2011 lies halfway between 2009 and 2013, so we must find the coordinates of the midpoint of the segment that has endpoints (2009, 124.0) and (2013, 137.4)

2009 2013 124.0 137.4,

2 2

2011, 130.7

M

The estimate of $130.7 billion is $0.1 billion more than the actual amount.


Choose any real number for x, substitute the value in the equation and then solve for y. Note that additional answers are possible.

(a) x y = −2x + 5

−1 2( 1) 5 7y

0 2(0) 5 5y

3 2(3) 5 1y

Three ordered pairs that are solutions are (−1, 7), (0, 5), and (3, −1). Other answers are possible.


Copyright © 2017 Pearson Education, Inc

(b) y 3 1x y

−9 3 39 1 8 2x

−2 3 32 1 1 1x

−1 3 31 1 0 0x

0 33 0 1 1 1x

7 3 37 1 8 2x

Ordered pairs that are solutions are (−2, −9), (−1, −2), (0, −1), (1, 0) and (2, 7). Other answers are possible.

(c) x 2 1y x

−2 2( 2) 1 3y

−1 2( 1) 1 0y

0 20 1 1y

1 21 1 0y

2 22 1 3y

Ordered pairs that are solutions are (−2, −3), (−1, 0), (0, 1), (1, 0) and (2, −3). Other answers are possible.


(a) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:

52

0 2 52(0) 5 5x x

y y

Find a third point on the graph by letting x = −1 and solving for y: 2 1 5 7y .

The three points are 52

, 0 , (0, 5), and (−1, 7).

Note that (3, −1) is also on the graph.

(b) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:

33

3

0 1 1 1

0 1 0 1 1

x

y y y

Find a third point by letting x = 2 and solving

for y: 332 1 2 1 7y y y .

Find a fourth point by letting x = −2 and solving for y:

332 1 2 1 9y y y

The points to be plotted are (0, −1), (1, 0), (2, 7), and (−2, −9). Note that (−1, −2) is also on the graph.

(c) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:

2 2 2

2

0 1 1 1 1

0 1 1

x x x x

y


for y: 22 1 3y . Find a fourth point

by letting x = −2 and solving for y:

22 1 3y

The points to be plotted are (−1, 0), (1, 0), (0, 1), (2, −3), and (−2, −3)

Section 2.2 Circles Classroom Example 1 (page 195)

(a) (h, k) = (1, −2) and r = 3

2 2 2

22 2

2 2

1 2 3

1 2 9

x h y k r

x y

x y

(b) (h, k) = (0, 0) and r = 2

2 2 2

2 2 2

2 2

0 0 2

4

x h y k r

x y

x y




(a) 2 21 2 9x y

This is a circle with center (1, −2) and radius 3.

(b) 2 2 4x y

This is a circle with center (0, 0) and radius 2.


Complete the square twice, once for x and once for y:

2 2

2 2

2 2

4 8 44 0

4 4 8 16 44 4 16

2 4 64

x x y y

x x y y

x y

Because c = 64 and 64 > 0, the graph is a circle. The center is (−2, 4) and the radius is 8.


2 22 2 2 6 45x y x y

Group the terms, factor out 2, and then complete the square:

2 21 92 2 3

4 4

1 945 2 2

4 4

x x y y

Factor and then divide both sides by 2:

2 2

2 2

1 32 2 50

2 2

1 325

2 2

x y

x y

Because c = 25 and 25 > 0, the graph is a circle. The

center is 1 3

,2 2

and the radius is 5.



2 2

2 2

2 2

6 2 12 0

6 9 2 1 12 9 1

3 1 2

x x y y

x x y y

x y

Because c = −2 and −2 < 0, the graph is nonexistent.


Determine the equation for each circle and then substitute x = −3 and y = 4. Station A:

2 2 2

2 2 2

2 2

1 4 4

3 1 4 4 4

4 416 16

x y

Station B:

2 2 2

2 2

2 2

2 2

6 0 5

6 25

3 6 4 25

3 4 259 16 25

25 25

x y

x y

Station C:

22 2

2 2

2 2

2 2

5 2 10

5 2 100

3 5 4 2 100

8 6 10064 36 100

100 100

x y

x y

Because (−3, 4) satisfies all three equations, we can conclude that the epicenter is (−3, 4).



Section 2.3 Functions Classroom Example 1 (page 204)

4,0 , 3,1 , 3,1M

M is a function because each distinct x value has exactly one y value.

2,3 , 3, 2 , 4,5 , 5, 4N

N is a function because each distinct x value has exactly one y value.

4,3 , 0,6 , 2,8 , 4, 3P

P is not a function because there are two y-values for x = −4.


(a) Domain: {−4, −1, 1, 3} Range: {−2, 0, 2, 5} The relation is a function.

(b) Domain: {1, 2, 3} Range: {4, 5, 6, 7} The relation is not a function because 2 maps to 5 and 6.

(c) Domain: {−3, 0, 3, 5} Range: {5} The relation is a function.


(a) Domain: {−2, 4}; range: {0, 3}

(b) Domain: ( , ) ; range: ( , )

(c) Domain: [−5, 5]; range: [−3, 3]

(d) Domain: ( , ) ; range: ( , 4]


(a)

Not a function

(b)

Function

(c)

Not a function

(d)

Function


(a) 2 5y x represents a function because y is

always found by multiplying x by 2 and subtracting 5. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers or ( , ) .

Because y is twice x, less 5, y also may be any real number, and so the range is also all real numbers, ( , ) .



(b) For any choice of x in the domain of 2 3y x , there is exactly one corresponding

value for y, so the equation defines a function. The function is defined for all values of x, so the domain is ( , ) . The square of any

number is always positive, so the range is [3, ) .

(c) For any choice of x in the domain of x y ,

there are two possible values for y. Thus, the equation does not define a function. The domain is [0, ) while the range is ( , ) .

(d) By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into y x corresponds to many values of y. The

ordered pairs (0, 2) (1, 1) (1, 0) (3, −1) and so on, all satisfy the inequality. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers, ( , ) .

(e) For 3

2y

x

, we see that y can be found by

dividing x + 2 into 3. This process produces one value of y for each value of x in the domain. The domain includes all real numbers except those that make the denominator equal to zero, namely x = −2. Therefore, the domain is ( , 2) ( 2, ) . Values of y can be

negative or positive, but never zero. Therefore the range is ( , 0) (0, ) .


(a) 2( 3) ( 3) 6( 3) 4 13f

(b) 2( ) 6 4f r r r

(c) ( 2) 3( 2) 1 3 7g r r r


(a) 2( 1) 2( 1) 9 7f

(b) ( 1) 6f

(c) ( 1) 5f

(d) ( 1) 0f


(a) 2

2

2

( ) 2 3

( 5) ( 5) 2( 5) 3 12

( ) 2 3

f x x x

f

f t t t

(b) 2

2 3 6 23

2( ) 2

32 16

( 5) ( 5) 23 3

2( ) 2

3

x y y x

f x x

f

f t t




The function is increasing on ( , 1), decreasing

on (–1, 1) and constant on (1, ).


The example refers to the following figure.

(a) The water level is changing most rapidly from 0 to 25 hours.

(b) The water level starts to decrease after 50 hours.

(c) After 75 hours, there are 2000 gallons of water in the pool.

Section 2.4 Linear Functions Classroom Example 1 (page 220)

32

( ) 6f x x ; Use the intercepts to graph the

function. 32


3 32 2

0 6 6 4 :x x x x-intercept

Domain: ( , ), range: ( , )


( ) 2f x is a constant function. Its graph is a

horizontal line with a y-intercept of 2.

Domain: ( , ), range: {2}


x = 5 is a vertical line intersecting the x-axis at (5, 0).

Domain:{5}, range: ( , )


3 4 0;x y Use the intercepts.

3 0 4 0 4 0 0 : -intercepty y y y

3 4 0 0 3 0 0 : -interceptx x x x


3 4 4 0 12 4 04 12 3

y yy y

Graph the line through (0, 0) and (4, −3).





(a) 4 ( 6) 10 5

2 2 4 2m

(b) 8 8 0

05 ( 3) 8

m

(c) 10 10 20

4 ( 4) 0m

the slope is

undefined.


2x − 5y = 10 Solve the equation for y. 2 5 10

5 2 102

25

x yy x

y x

The slope is 2

,5

the coefficient of x.


First locate the point (−2, −3). Because the slope is 43

, a change of 3 units horizontally (3 units to the

right) produces a change of 4 units vertically (4 units up). This gives a second point, (1, 1), which can be used to complete the graph.


The average rate of change per year is 8658 9547 889

296.33 million2013 2010 3

The graph confirms that the line through the ordered pairs fall from left to right, and therefore has negative slope. Thus, the amount spent by the federal government on R&D for general science decreased by an average of $296.33 million (or $296,330,000) each year from 2010 to 2013.


(a) ( ) 120 2400C x x

(b) ( ) 150R x x

(c) ( ) ( ) ( )150 (120 2400)30 2400

P x R x C xx x

x

(d) ( ) 0 30 2400 0 80P x x x

At least 81 items must be sold to make a profit.



( 5) 2( 3)5 2 6

2 1

y xy x

y x


First find the slope: 3 ( 1) 4

4 5 9m

Now use either point for 1 1( , )x y : 49

3 [ ( 4)]9( 3) 4( 4)9 27 4 16

9 4 114 9 11

y xy xy x

y xx y


Write the equation in slope-intercept form: 34

3 4 12 4 3 12 3x y y x y x

The slope is 34

, and the y-intercept is (0, −3).




First find the slope: 4 2 1

2 2 2m

Now, substitute 12

for m and the coordinates of one

of the points (say, (2, 2)) for x and y into the slope-intercept form y = mx + b, then solve for b:

12

2 2 3b b . The equation is

12

3y x .


The example refers to the following figure:

(a) The line rises 5 units each time the x-value

increases by 2 units. So the slope is 52

. The

y-intercept is (0, 5), and the x-intercept is (−2, 0).


( ) 5f x x .


(a) Rewrite the equation 3 2 5x y in slope-

intercept form to find the slope: 3 52 2

3 2 5x y y x The slope is 32

.

The line parallel to the equation also has slope 32

. An equation of the line through (2, −4) that

is parallel to 3 2 5x y is 3 32 2

( 4) ( 2) 4 3y x y x

32

7y x or 3x − 2y = 14.

(b) The line perpendicular to the equation has

slope 23

. An equation of the line through

(2, −4) that is perpendicular to 3 2 5x y is 2 2 43 3 3

( 4) ( 2) 4y x y x 82

3 3y x or 2x +3y = −8.


(a) First find the slope: 7703 6695

5043 1

m

Now use either point for 1 1( , )x y :

6695 504( 1)6695 504 504

504 6191

y xy x

y x

(b) The year 2015 is represented by x = 6. 504(6) 6191 9215y

According to the model, average tuition and fees for 4-year colleges in 2015 will be about $9215.


Write the equation as an equivalent equation with 0 on one side.

3 2(5 ) 2 38x x x

3 2(5 ) 2 38 0x x x

Now graph the equation to find the x-intercept.

The solution set is {−4}.

Section 2.6 Graphs of Basic Functions Classroom Example 1 (page 249)

(a) The function is continuous over ( , 0) (0, )

(b) The function is continuous over its entire domain ( , ).




(a) Graph each interval of the domain separately. If x < 1, the graph of f(x) = 2x + 4 has an endpoint at (1, 6), which is not included as part of the graph. To find another point on this part of the graph, choose x = 0, so y = 4. Draw the ray starting at (1, 6) and extending through (0, 4). Graph the function for x ≥ 1, f(x) = 4 − x similarly. This part of the graph has an endpoint at (1, 3), which is included as part of the graph. Find another point, say (4, 0), and draw the ray starting at (1, 3) which extends through (4, 0).

(b) Graph each interval of the domain separately. If x ≤ 0, the graph of f(x) = −x − 2 has an endpoint at (0, −2), which is included as part of the graph. To find another point on this part of the graph, choose x = −2, so y = 0. Draw the ray starting at (0, −2) and extending through (−2, 0). Graph the function for x > 0,

2( ) 2f x x similarly. This part of the graph

has an endpoint at (0, −2), which is not included as part of the graph. Find another point, say (2, 2), and draw the curve starting at (0, −2) which extends through (2, 2). Note that the two endpoints coincide, so (0, −2) is included as part of the graph.


Create a table of sample ordered pairs:

x −6 −3 32

0 32

3 6

13

2y x −4 −3 −3 −2 −2 −1 0


For x in the interval (0, 2], y = 20. For x in (2, 3], y = 20 + 2 = 22. For x in (3, 4], y = 22 + 2 = 24. For x in (4, 5], y = 24 + 2 = 26. For x in (5, 6], y = 26 + 2 = 28.

Section 2.7 Graphing Techniques Classroom Example 1 (page 260)

Use this table of values for parts (a)−(c)

x 2( ) 2g x x 212

( )h x x 212

( )k x x

−2 8 2 1

−1 2 12

14

0 0 0 0

1 2 12

14

2 8 2 1

(a) 2( ) 2g x x



(b) 212

( )h x x

(c) 2

1( )

2k x x


Use this table of values for parts (a) and (b)

x ( )g x x ( )h x x

−2 −2 2

−1 −1 1

0 0 0

1 −1 1

2 −2 2

(a) ( )g x x

(b) ( )h x x


(a) x y

Replace x with –x to obtain x y . The result is not the same as the

original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain x y x y The

result is the same as the original equation, so the graph is symmetric with respect to the x-axis. The graph is symmetric with respect to the x-axis only.

(b) 3y x

Replace x with –x to obtain 3y x

3y x . The result is the same as the

original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain 3y x . The result is not the same

as the original equation, so the graph is not symmetric with respect to the x-axis. Therefore, the graph is symmetric with respect to the y-axis only.

(c) 2 6x y


2 6x y . The result is not the same as the

original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain 2 ( ) 6x y


original equation, so the graph is not symmetric with respect to the x-axis. Therefore, the graph is not symmetric with respect to either axis.

(d) 2 2 25x y

Replace x with –x to obtain 2 2( ) 25x y 2 2 25x y . The result is

the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain

2 2( ) 25x y 2 2 25x y . The result is

the same as the original equation, so the graph is symmetric with respect to the x-axis. Therefore, the graph is symmetric with respect to both axes. Note that the graph is a circle of radius 5, centered at the origin.




(a) 32y x

Replace x with –x and y with –y to obtain 3 3 3( ) 2( ) 2 2y x y x y x . The

result is the same as the original equation, so the graph is symmetric with respect to the origin.

(b) 22y x

Replace x with –x and y with –y to obtain 2 2 2( ) 2( ) 2 2y x y x y x .

The result is not the same as the original equation, so the graph is not symmetric with respect to the origin.


(a) 5 3( ) 2 3g x x x x

Replace x with –x to obtain 5 3

5 3

5 3

( ) ( ) 2( ) 3( )

2 3

( 2 3 ) ( )

g x x x x

x x x

x x x g x

g(x) is an odd function.

(b) 2( ) 2 3h x x

Replace x with –x to obtain 2 2( ) 2( ) 3 2 3 ( )h x x x h x h(x) is

an even function.

(c) 2( ) 6 9k x x x

Replace x with –x to obtain 2

2( ) ( ) 6( ) 9

6 9 ( ) and ( )

k x x x

x x k x k x

k(x) is neither even nor odd.


Compare a table of values for 2( )g x x with 2( ) 2f x x . The graph of f(x) is the same as the

graph of g(x) translated 2 units up.

x 2( )g x x 2( ) 2f x x

−2 4 6

−1 1 3

0 0 2

1 1 3

2 4 6


Compare a table of values for 2( )g x x with 2( ) ( 4)f x x . The graph of f(x) is the same as the

graph of g(x) translated 4 units left.

x 2( )g x x 2( ) ( 4)f x x

−7 49 9

−6 36 4

−5 25 1

−4 16 0

−3 9 1

−2 4 4

−1 1 9


(a) 2( ) ( 1) 4f x x

This is the graph of 2( )g x x , translated one

unit to the right, reflected across the x-axis, and then translated four units up.

x 2( )g x x 2( ) ( 1) 4f x x

−2 4 −5

−1 1 0

0 0 3

1 1 4

2 4 3

3 9 0

4 16 −5




(continued)

(b) ( ) 2 3f x x

This is the graph of ( )g x x , translated three

units to the left, reflected across the x-axis, and then stretched vertically by a factor of two.

x ( )g x x ( ) 2 3f x x

−6 6 −6

−5 5 −4

−4 4 −2

−3 3 0

−2 2 −2

−1 1 −4

0 0 −6

(c) 12

( ) 2 3h x x

This is the graph of ( )g x x , translated two

units to the left, shrunk vertically by a factor of 2, and then translated 3 units down.

x ( )g x x 12

( ) 2 3h x x

−2 undefined −3

−1 undefined −2.5

0 0 12

2 3 2.3

2 2 1.4 −2

6 6 2.4 12

8 3 1.6

7 7 2.6 −1.5


The graphs in the exercises are based on the following graph.

(a) ( ) ( ) 2g x f x

This is the graph of f(x) translated two units down.

(b) ( ) ( 2)h x f x

This is the graph of f(x) translated two units right.



(c) ( ) ( 1) 2k x f x

This is the graph of f(x) translated one unit left, and then translated two units up.

(d) ( ) ( )F x f x

This is the graph of f(x) reflected across the y-axis.



For parts (a)−(d), ( ) 3 4f x x and 2( ) 2 1g x x

(a) (0) 3(0) 4 4f and 2(0) 2(0) 1 1g , so

( )(0) 4 1 5f g

(b) (4) 3(4) 4 8f and 2(4) 2(4) 1 31g ,

so ( )(4) 8 31 23f g

(c) ( 2) 3( 2) 4 10f and 2( 2) 2( 2) 1 7g , so

( )( 2) ( 10)(7) 70fg

(d) (3) 3(3) 4 5f and 2(3) 2(3) 1 17g ,

so 5

(3)17

f

g


For parts (a)−(e), 2( ) 3f x x x and ( ) 4 5g x x

(a) 2 2( )( ) ( 3 ) (4 5) 5f g x x x x x x

(b) 2 2( )( ) ( 3 ) (4 5) 7 5f g x x x x x x

(c) 2

3 2 2

3 2

( )( ) ( 3 )(4 5)

4 5 12 15

4 7 15

fg x x x x

x x x x

x x x

(d) 2 3

( )4 5

f x xx

g x

(e) The domains of f and g are both ( , ) . So,

the domains of f + g, f − g, and fg are the intersection of the domains of f and g,

( , ) . The domain of fg

includes those

real numbers in the intersection of the domains of f and g for which ( ) 4 5 0g x x

54

x . So the domain of fg

is

5 54 4

, , .


(a)

From the figure, we have f(1) = 3 and

g(1) = 1, so (f + g)(1) = 3 + 1 = 4. f(0) = 4 and g(0) = 0, so (f − g)(0) = 4 − 0 = 4. f(−1) = 3 and g(−1) = 1, so (f g)(−1) = (3)(1) = 3

f(−2) = 0 and g(−2) = 2, so 02

( 2) 0fg

.

(b) x f(x) g(x)

−2 −5 0

−1 −3 2

0 −1 4

1 1 6

From the table, we have f(1) = 1 and g(1) = 6, so (f + g)(1) = 1 + 6 = 7. f(0) = −1 and g(0) = 4, so (f − g)(1) = −1 − 4 = −5. f(−1) = −3 and g(−1) = 2, so (f g)(−1) = (−3)(2) = −6 f(−2) = −5 and g(−2) = 0, so

50

( 2)fg

fg

is undefined.



(c) ( ) 3 4, ( )f x x g x x

From the formulas, we have (1) 3(1) 4 7f and (1) 1 1g , so

(f + g)(1) = 7 + (−1) = 6. (0) 3(0) 4 4f and (0) 0 0g , so

(f − g)(1) = 4 − 0 = 4. ( 1) 3( 1) 4 1f and ( 1) 1 1g ,

so (fg)(−1) =(1)(−1) = −1. ( 2) 3( 2) 4 2f and

( 2) 2 2g , so 2

( 2) 1.2

f

g


Step 1: Find ( )f x h : 2

2 2

2 2

( ) 3( ) 2( ) 4

3( 2 ) 2 2 4

3 6 3 2 2 4

f x h x h x h

x xh h x h

x xh h x h

Step 2: Find ( ) ( )f x h f x :

2 2 2

2

( ) ( )

(3 6 3 2 2 4) (3 2 4)

6 3 2

f x h f x

x xh h x h x x

xh h h

Step 3: Find the difference quotient: 2( ) ( ) 6 3 2

6 3 2f x h f x xh h h

x hh h


For parts (a) and (b), ( ) 4f x x and 2

( )g xx

(a) First find g(2): 2

(2) 12

g . Now find

( )(2) ( (2)) (1) 1 4 5f g f g f

(b) First find f(5): (5) 5 4 9 3f . Now

find 23

( )(5) ( (5)) (3)g f g f g

The screens show how a graphing calculator

evaluates the expressions in this classroom example.


For parts (a) and (b), ( ) 1f x x and

( ) 2 5g x x

(a) ( )( ) ( ( )) (2 5) 1 2 4f g x f g x x x

The domain and range of g are both ( , ) .

However, the domain of f is [1, ) . Therefore,

( )g x must be greater than or equal to 1:

2 5 1 2x x . So, the domain of f g

is [ 2, ) .

(b) ( )( ) ( ( )) 2 1 5g f x g f x x

The domain of f is [1, ) , while the range of f

is [0, ) . The domain of g is ( , ) .

Therefore, the domain of ( )g f is restricted

to that portion of the domain of g that intersects with the domain of f, that is [1, ) .


For parts (a) and (b), 5

( )4

f xx

and 2

( )g xx

(a) 5 5

( )( ) ( ( ))2 4 2 4

xf g x f g x

x x

The domain and range of g are both all real numbers except 0. The domain of f is all real numbers except −4. Therefore, the expression for ( )g x cannot equal −4. So,

2 14

2x

x . So, the domain of f g

is the set of all real numbers except for 12

,

and 0. This is written

1 12 2

, , 0 0,



(b) 2 2 8

( )( ) ( ( ))5 ( 4) 5

xg f x g f x

x

The domain of f is all real numbers except −4, while the range of f is all real numbers except 0. The domain and range of g are both all real numbers except 0, which is not in the range of f. So, the domain of g f is the set of all real

numbers except for −4. This is written , 4 4,


( ) 2 5f x x and 2( ) 3g x x x 2

2

2

2 2

2

( )( ) (2 5) 3(2 5) (2 5)

3(4 20 25) 2 5

12 58 70

( )( ) (3 ) 2(3 ) 5

6 2 5

g f x g x x x

x x x

x x

f g x f x x x x

x x

In general, 2 212 58 70 6 2 5,x x x x so

( )( ) ( )( )g f x f g x .


2( )( ) 4(3 2) 5(3 2) 8f g x x x

Answers may vary. Sample answer: 2( ) 4 5 8f x x x and ( ) 3 2g x x .

Copyright © 2017 Pearson Education, Inc. 170

Chapter 2 GRAPHS AND FUNCTIONS

Section 2.1 Rectangular Coordinates and Graphs

1. The point (–1, 3) lies in quadrant II in the rectangular coordinate system.

2. The point (4, 6) lies on the graph of the equation y = 3x – 6. Find the y-value by letting x = 4 and solving for y.

3 4 6 12 6 6y

3. Any point that lies on the x-axis has y-coordinate equal to 0.

4. The y-intercept of the graph of y = –2x + 6 is (0, 6).

5. The x-intercept of the graph of 2x + 5y = 10 is (5, 0). Find the x-intercept by letting y = 0 and solving for x.

2 5 0 10 2 10 5x x x

6. The distance from the origin to the point (–3, 4) is 5. Using the distance formula, we have

2 2

2 2

( , ) ( 3 0) (4 0)

3 4 9 16 25 5

d P Q

7. True

8. True

9. False. The midpoint of the segment joining (0, 0) and (4, 4) is

4 0 4 0 4 4, , 2, 2 .

2 2 2 2

10. False. The distance between the point (0, 0) and (4, 4) is

2 2 2 2( , ) (4 0) (4 0) 4 4

16 16 32 4 2

d P Q

11. Any three of the following: 2, 5 , 1,7 , 3, 9 , 5, 17 , 6, 21

12. Any three of the following: 3,3 , 5, 21 , 8,18 , 4,6 , 0, 6

13. Any three of the following: (1999, 35), (2001, 29), (2003, 22), (2005, 23), (2007, 20), (2009, 20)

14. Any three of the following:

2002,86.8 , 2004, 89.8 , 2006, 90.7 ,

2008, 97.4 , 2010, 106.5 , 2012,111.4 ,

2014, 111.5

15. P(–5, –6), Q(7, –1)

(a) 2 2

2 2

( , ) [7 – (–5)] [ 1 – (–6)]

12 5 169 13

d P Q


–5 7 6 ( 1) 2 7, ,

2 2 2 27

1, .2



16. P(–4, 3), Q(2, –5)

(a) 2 2

2 2

( , ) [2 – (– 4)] (–5 – 3)

6 (–8) 100 10

d P Q


– 4 2 3 (–5) 2 2, ,

2 2 2 21, 1 .

17. (8, 2), (3, 5)P Q

(a)

2 2

2 2

( , ) (3 – 8) (5 – 2)

5 3

25 9 34

d P Q


8 3 2 5 11 7, ,

2 2 2 2

.

18. P (−8, 4), Q (3, −5)

(a) 2 2

2 2

( , ) 3 – 8 5 4

11 (–9) 121 81

202

d P Q


–8 3 4 (–5) 5 1, , .

2 2 2 2

19. P(–6, –5), Q(6, 10)

(a) 2 2

2 2

( , ) [6 – (– 6)] [10 – (–5)]

12 15 144 225

369 3 41

d P Q


– 6 6 –5 10 0 5 5, , 0, .

2 2 2 2 2

20. P(6, –2), Q(4, 6)

(a) 2 2

2 2

( , ) (4 – 6) [6 – (–2)]

(–2) 8

4 64 68 2 17

d P Q


6 4 2 6 10 4, , 5, 2

2 2 2 2

21. 3 2, 4 5 ,P 2, – 5Q

(a)

2 2

2 2

( , )

2 – 3 2 – 5 – 4 5

–2 2 –5 5

8 125 133

d P Q


3 2 2 4 5 (– 5),

2 2

4 2 3 5 3 5, 2 2, .

2 2 2

22. – 7, 8 3 , 5 7, – 3P Q

(a) 2 2

2 2

( , )

[5 7 – (– 7)] (– 3 – 8 3)

(6 7) (–9 3) 252 243

495 3 55

d P Q


– 7 5 7 8 3 (– 3),

2 2

4 7 7 3 7 3, 2 7, .

2 2 2

23. Label the points A(–6, –4), B(0, –2), and C(–10, 8). Use the distance formula to find the length of each side of the triangle.

2 2

2 2

( , ) [0 – (– 6)] [–2 – (– 4)]

6 2 36 4 40

d A B

2 2

2 2

( , ) (–10 – 0) [8 – (–2)]

( 10) 10 100 100

200

d B C

2 2

2 2

( , ) [–10 – (– 6)] [8 – (– 4)]

(– 4) 12 16 144 160

d A C

Because 2 2 240 160 200 ,




24. Label the points A(–2, –8), B(0, –4), and C(–4, –7). Use the distance formula to find the length of each side of the triangle.

2 2

2 2

( , ) [0 – (–2)] [– 4 – (–8)]

2 4 4 16 20

d A B

2 2

2 2

( , ) (– 4 – 0) [–7 – (– 4)]

(– 4) (–3) 16 9

25 5

d B C

2 2

2 2

( , ) [– 4 – (–2)] [–7 – (–8)]

(–2) 1 4 1 5

d A C

Because 2 2 2( 5) ( 20) 5 20 25 5 , triangle ABC is a right triangle.


2 2

2 2

( , ) [1 – (– 4)] (4 – 1)

5 3 25 9 34

d A B

2 2

2 2

( , ) (– 6 –1) (–1 – 4)

(–7) (–5) 49 25 74

d B C

2 2

2 2

( , ) [– 6 – (– 4)] (–1 –1)

(–2) (–2) 4 4 8

d A C

Because 2 2 2( 8) ( 34) ( 74) because

8 34 42 74, triangle ABC is not a right triangle.

26. Label the points A(–2, –5), B(1, 7), and C(3, 15).

2 2

2 2

( , ) [1 ( 2)] [7 ( 5)]

3 12 9 144 153

d A B

2 2

2 2

( , ) (3 1) (15 7)

2 8 4 64 68

d B C

2 2

2 2

( , ) [3 ( 2)] [15 ( 5)]

5 20 25 400 425

d A C

Because 22 2( 68) ( 153) 425 because

68 153 221 425 , triangle ABC is not a right triangle.


2 2

2 2

( , ) 2 – –4 5 3

6 2 36 4 40

d A B

2 2

2 2

( , ) 1 2 6 5

(– 3) (–11)

9 121 130

d B C

2 2

22

( , ) –1 – –4 6 3

3 9 9 81 90

d A C


ABC is a right triangle.

28. Label the points A(–7, 4), B(6, –2), and C(0, –15).

2 2

22

( , ) 6 – –7 2 4

13 6

169 36 205

d A B

22

2 2

( , ) 0 6 –15 – –2

– 6 –13

36 169 205

d B C

2 2

22

( , ) 0 – –7 15 4

7 19 49 361 410

d A C

Because 2 2 2205 205 410 ,


29. Label the given points A(0, –7), B(–3, 5), and C(2, –15). Find the distance between each pair of points.

22

2 2

( , ) 3 0 5 – –7

–3 12 9 144

153 3 17

d A B

2 2

22

( , ) 2 – 3 –15 – 5

5 –20 25 400

425 5 17

d B C

22

22

( , ) 2 0 15 – 7

2 –8 68 2 17

d A C

Because ( , ) ( , ) ( , )d A B d A C d B C or

3 17 2 17 5 17, the points are collinear.



30. Label the points A(–1, 4), B(–2, –1), and C(1, 14). Apply the distance formula to each pair of points.

2 2

2 2

( , ) –2 – –1 –1 – 4

–1 –5 26

d A B

2 2

2 2

( , ) 1 – –2 14 – –1

3 15 234 3 26

d B C

2 2

2 2

( , ) 1 – –1 14 – 4

2 10 104 2 26

d A C

Because 26 2 26 3 26 , the points are collinear.

31. Label the points A(0, 9), B(–3, –7), and C(2, 19).

2 2

2 2

( , ) (–3 – 0) (–7 – 9)

(–3) (–16) 9 256

265 16.279

d A B

2 2

2 2

( , ) 2 – –3 19 – –7

5 26 25 676

701 26.476

d B C

2 2

2 2

( , ) 2 – 0 19 – 9

2 10 4 100

104 10.198

d A C

Because ( , ) ( , ) ( , )d A B d A C d B C

or 265 104 70116.279 10.198 26.476,

26.477 26.476,


32. Label the points A(–1, –3), B(–5, 12), and C(1, –11).

2 2

2 2

( , ) –5 – –1 12 – –3

– 4 15 16 225

241 15.5242

d A B

2 2

22

( , ) 1 – –5 –11 –12

6 –23 36 529

565 23.7697

d B C

2 2

22

( , ) 1 – –1 –11 – –3

2 –8 4 64

68 8.2462

d A C

Because d(A, B) + d(A, C) d(B, C)

or 241 68 56515.5242 8.2462 23.7697

23.7704 23.7697,


33. Label the points A(–7, 4), B(6,–2), and C(–1,1).

2 2

22

( , ) 6 – –7 2 4

13 6 169 36

205 14.3178

d A B

22

2 2

( , ) 1 6 1 –2

7 3 49 9

58 7.6158

d B C

2 2

22

( , ) 1 – –7 1 4

6 –3 36 9

45 6.7082

d A C

Because d(B, C) + d(A, C) d(A, B) or

58 45 2057.6158 6.7082 14.3178

14.3240 14.3178,


34. Label the given points A(–4, 3), B(2, 5), and C(–1, 4). Find the distance between each pair of points.

2 2 2 2( , ) 2 – –4 5 3 6 2

36 4 40 2 10

d A B

2 2

2 2

( , ) ( 1 2) (4 5)

3 (–1) 9 1 10

d B C

2 2

2 2

( , ) 1 – 4 4 3

3 1 9 1 10

d A C

Because ( , ) ( , ) ( , )d B C d A C d A B or

10 10 2 10, the points are collinear.




5 and 82 2

13 10 and 10 16

–3 and 6.

x y

x yx y



–7 and 62 2

–9 –14 and 9 12

–5 and 3.

x y

x yx y



12 and 62 2

19 24 and 16 12

5 and – 4.

x y

x yx y

The other endpoint has coordinates (5, –4).


–9 and 82 2

–16 –18 and 9 16

–2 and 7

x y

x yx y


39. Midpoint (a, b), endpoint (p, q)

and2 2

2 and 2

2 and 2

p x q ya b

p x a q y bx a p y b q

The other endpoint has coordinates (2 , 2 )a p b q .

40. Midpoint 6 , 6a b , endpoint 3 , 5a b

3 56 and 6

2 23 12 and 5 12

9 and 7

a x b ya b

a x a b y bx a y b

The other endpoint has coordinates (9a, 7b).

41. The endpoints of the segment are (1990, 21.3) and (2012, 30.1).

1990 2012 21.3 30.9,

2 22001, 26.1

M

The estimate is 26.1%. This is very close to the actual figure of 26.2%.

42. The endpoints are (2006, 7505) and (2012, 3335)

2006 2012 7505 3335,

2 22009, 5420

M

According to the model, the average national advertising revenue in 2009 was $5420 million. This is higher than the actual value of $4424 million.

43. The points to use are (2011, 23021) and (2013, 23834). Their midpoint is

2011 2013 23,021 23,834,

2 2(2012, 23427.5).

In 2012, the poverty level cutoff was approximately $23,428.

44. (a) To estimate the enrollment for 2003, use the points (2000, 11,753) and (2006, 13,180)

2000 2006 11,753 13,180,

2 22003, 12466.5

M

The enrollment for 2003 was about 12,466.5 thousand.

(b) To estimate the enrollment for 2009, use the points (2006, 13,180) and (2012, 14,880)

2006 2012 13,180 14,880,

2 22009, 14030

M

The enrollment for 2009 was about 14,030 thousand.


1 2 1 2, .2 2

x x y y

2 21 2 1 2

1 1

2 21 2 1 1 2 1

2 22 1 2 1

2 22 1 2 1

2 22 1 2 1

2 212 1 2 12

( , )

– –2 2

2 2– –

2 2 2 2

2 2

4 4

4

d P M

x x y yx y

x x x y y y

x x y y

x x y y

x x y y

x x y y




(continued)

2 21 2 1 2

2 2

2 22 1 2 2 1 2

2 22 1 2 1

2 22 1 2 1

2 22 1 2 1

2 212 1 2 12

( , )

2 2

2 2

2 2 2 2

2 2

4 4

4

d M Q

x x y yx y

x x x y y y

x x y y

x x y y

x x y y

x x y y

2 22 1 2 1( , )d P Q x x y y

Because

2 212 1 2 12

2 212 1 2 12

2 22 1 2 1 ,

x x y y

x x y y

x x y y

this shows ( , ) ( , ) ( , )d P M d M Q d P Q and

( , ) ( , ).d P M d M Q

46. The distance formula, 2 2

2 1 2 1( – ) ( – ) ,d x x y y can be written

as 2 2 1/ 22 1 2 1[( – ) ( – ) ] .d x x y y

In exercises 47−58, other ordered pairs are possible.


12

00 2 2

xy

4 0 x-intercept:

1212

00 2

2 4

yxx x


(b)


0

10 2 2

2

x

y

4 0 x-intercept: 0

10 2

21

2 42

y

x

x x


(b)

49. (a) x y 0 5

3 y-intercept:

53

02 0 3 5

3 5

xy

y y

52

0 x-intercept:

52

02 3 0 5

2 5

yxx x


(b)




0

3 0 2 62 6 3

xy

y y

2 0 x-intercept:

0

3 2 0 63 6 2

yxx x


(b)


20 0



(b)


2

0

0 20 2 2

xyy y

−1 3 additional point 2 6 additional point

no x-intercept: 2

20 0 2

2 2

y xx x

(b)

53. (a) x y 3 0 x-intercept:

0

0 30 3 3

yx

x x

4 1 additional point 7 2 additional point

no y-intercept:

0 0 3 3x y y

(b)


0

0 30 3 3

xyy y


9 0 x-intercept: 0

0 3

3 9

yxx x

(b)




00 2

2 2

xyy y

2 0 x-intercept: 0

0 20 2 2

yxx x



(b)

56. (a) x y −2 −2 additional point −4 0 x-intercept:

00 4

0 40 4 4

yx

xx x

0 −4 y-intercept: 0

0 4

4 4

xyy y

(b)


30 0 −1 −1 additional point 2 8 additional point

(b)


30 0 1 −1 additional point2 −8 additional point

(b)

59. Points on the x-axis have y-coordinates equal to 0. The point on the x-axis will have the same x-coordinate as point (4, 3). Therefore, the line will intersect the x-axis at (4, 0).

60. Points on the y-axis have x-coordinates equal to 0. The point on the y-axis will have the same y-coordinate as point (4, 3). Therefore, the line will intersect the y-axis at (0, 3).

61. Because (a, b) is in the second quadrant, a is negative and b is positive. Therefore, (a, – b) will have a negative x–coordinate and a negative y-coordinate and will lie in quadrant III. (–a, b) will have a positive x-coordinate and a positive y-coordinateand will lie in quadrant I. (–a, – b) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV. (b, a) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV.



62. Label the points ( 2, 2),A (13,10),B

(21, 5),C and (6, 13).D To determine which points form sides of the quadrilateral (as opposed to diagonals), plot the points.


each side.

2 2

2 2

2 2

22

( , ) 13 2 10 2

15 8 225 64

289 17

( , ) 21 13 5 10

8 15 64 225

289 17

d A B

d B C

22

2 2

( , ) 6 21 13 5

15 8

225 64 289 17

d C D

22

2 2

( , ) 2 6 2 13

8 15

64 225 289 17

d D A

Because all sides have equal length, the four points form a rhombus.

63. To determine which points form sides of the quadrilateral (as opposed to diagonals), plot the points.


each side.

2 2

2 2

( , ) 5 1 2 1

4 1 16 1 17

d A B

2 2

2 2

( , ) 3 5 4 2

2 2 4 4 8

d B C

2 2

2 2

( , ) 1 3 3 4

4 1

16 1 17

d C D

2 2

22

( , ) 1 1 1 3

2 2 4 4 8

d D A

Because d(A, B) = d(C, D) and d(B, C) = d(D, A), the points are the vertices of a parallelogram. Because d(A, B) ≠ d(B, C), the points are not the vertices of a rhombus.

64. For the points A(4, 5) and D(10, 14), the difference of the x-coordinates is 10 – 4 = 6 and the difference of the y-coordinates is 14 – 5 = 9. Dividing these differences by 3, we obtain 2 and 3, respectively. Adding 2 and 3 to the x and y coordinates of point A, respectively, we obtain B(4 + 2, 5 + 3) or B(6, 8).

Adding 2 and 3 to the x- and y- coordinates of point B, respectively, we obtain C(6 + 2, 8 + 3) or C(8, 11). The desired points are B(6, 8) and C(8, 11).

We check these by showing that d(A, B) = d(B, C) = d(C, D) and that d(A, D) = d(A, B) + d(B, C) + d(C, D).

2 2

2 2

2 2

2 2

( , ) 6 4 8 5

2 3 4 9 13

( , ) 8 6 11 8

2 3 4 9 13

d A B

d B C

2 2

2 2

2 2

2 2

( , ) 10 8 14 11

2 3 4 9 13

( , ) 10 4 14 5

6 9 36 81

117 9(13) 3 13

d C D

d A D

d(A, B), d(B, C), and d(C, D) all have the same measure and d(A, D) = d(A, B) + d(B, C) + d(C, D) Because

3 13 13 13 13.



Section 2.2 Circles

1. The circle with equation 2 2 49x y has center with coordinates (0, 0) and radius equal to 7.

2. The circle with center (3, 6) and radius 4 has

equation 23 6 16.x y

3. The graph of 2 24 7 9x y has center

with coordinates (4, –7).

4. The graph of 22 5 9x y has center with

coordinates (0, 5).

5. This circle has center (3, 2) and radius 5. This is graph B.

6. This circle has center (3, –2) and radius 5. This is graph C.

7. This circle has center (–3, 2) and radius 5. This is graph D.

8. This circle has center (–3, –2) and radius 5. This is graph A.

9. The graph of 2 2 0x y has center (0, 0) and radius 0. This is the point (0, 0). Therefore, there is one point on the graph.

10. 100 is not a real number, so there are no

points on the graph of 2 2 100.x y


2 2

2 2 2 2 2

0 0 6

0 0 6 36

x y

x y x y

(b)


2 2

2 2 2 2 2

0 0 9

0 0 9 81

x y

x y x y

(b)


2 2

2 2 2

2 2

2 0 6

2 0 6

( – 2) 36

x y

x yx y

(b)


2 2

2 2

3 0 3

3 9

x yx y

(b)


2 2

22

0 4 4

4 16

x yx y

(b)




22

22 2

2 2

0 3 7

0 3 7

( 3) 49

x y

x yx y

(b)

17. (a) Center (–2, 5), radius 4

2 2

2 2 2

2 2

2 5 4

[ – (–2)] ( – 5) 4

( 2) ( – 5) 16

x y

x yx y

(b)


2 2

2 2 2

2 2

4 3 5

4 3 5

4 3 25

x y

x yx y

(b)


22

2 2 2

2 2

5 4 7

( – 5) [ – (– 4)] 7

( – 5) ( 4) 49

x y

x yx y

(b)

20. (a) Center (–3, –2), radius 6

2 2

2 2 2

2 2

3 2 6

3 2 6

( 3) ( 2) 36

x y

x yx y

(b)


2 2

2 2

2 2 2

2 2 2

x y

x y

(b)


2 2

2 2 2

2 2

3 3 3

3 3 3

3 3 3

x y

x y

x y



(b)

23. (a) The center of the circle is located at the midpoint of the diameter determined by the points (1, 1) and (5, 1). Using the midpoint formula, we have

1 5 1 1, 3,1

2 2C

. The radius is

one-half the length of the diameter:

2 215 1 1 1 2

2r


2 23 1 4x y



2 2

2 2

2 2

3 1 46 9 2 1 4

6 2 6 0

x yx x y y

x y x y

24. (a) The center of the circle is located at the midpoint of the diameter determined by the points (−1, 1) and (−1, −5).

Using the midpoint formula, we have

1 ( 1) 1 ( 5), 1, 2 .

2 2C


2 211 1 5 1 3

2r


2 21 2 9x y



2 2

2 2

2 2

1 2 92 1 4 4 9

2 4 4 0

x yx x y y

x y x y

25. (a) The center of the circle is located at the midpoint of the diameter determined by the points (−2, 4) and (−2, 0). Using the midpoint formula, we have

2 ( 2) 4 0, 2, 2 .

2 2C


2 212 2 4 0 2

2r


2 22 2 4x y



2 2

2 2

2 2

2 2 44 4 4 4 4

4 4 4 0

x yx x y y

x y x y

26. (a) The center of the circle is located at the midpoint of the diameter determined by the points (0, −3) and (6, −3). Using the midpoint formula, we have

0 6 3 ( 3), 3, 3

2 2C

.


2216 0 3 3 3

2r


2 23 3 9x y



2 2

2 2

2 2

3 3 96 9 6 9 9

6 6 9 0

x yx x y y

x y x y


2 2

2 2

2 2

6 8 –9

6 9 8 16 –9 9 16

3 4 16

x x y y

x x y y

x y

Yes, it is a circle. The circle has its center at (–3, –4) and radius 4.




2 2

2 2

2 2

8 – 6 16

8 16 – 6 9 –16 16 9

4 – 3 9

x x y y

x x y y

x y

Yes, it is a circle. The circle has its center at (–4, 3) and radius 3.

29. 2 2 4 12 4x y x y Complete the square on x and y separately.

2 2

2 2

2 2

– 4 12 – 4

– 4 4 12 36 – 4 4 36

– 2 6 36

x x y y

x x y y

x y


30. 2 2 –12 10 –25x y x y Complete the square on x and y separately.

2 2

2 2

2 2

–12 10 –25

–12 36 10 25

– 25 36 25

– 6 5 36

x x y y

x x y y

x y


31. 2 24 4 4 –16 – 19 0x y x y Complete the square on x and y separately.

2 2

2 214

14

4 4 – 4 19

4 4 – 4 4

19 4 4 4

x x y y

x x y y

2 212

2 212

4 4 – 2 36

– 2 9

x y

x y


, 2 and

radius 3.

32. 2 29 9 12 – 18 – 23 0x y x y Complete the square on x and y separately.

2 243

2 24 43 9

49

9 9 – 2 23

9 9 – 2 1

23 9 9 1

x x y y

x x y y

2 223

2 223

9 9 –1 36

–1 4

x y

x y


, 1 and

radius 2.


2 2

2 2

2 2

2 – 6 –14

2 1 – 6 9 –14 1 9

1 – 3 – 4

x x y y

x x y y

x y



2 2

2 2

2 2

4 – 8 –32

4 4 – 8 16

– 32 4 16

2 – 4 –12

x x y y

x x y y

x y



2 2

2 2

2 2

6 6 18

6 9 6 9 18 9 9

3 3 0

x x y y

x x y y

x y

The graph is the point (3, 3).


2 2

2 2

2 2

4 4 8

4 4 4 4 8 4 4

2 2 0

x x y y

x x y y

x y

The graph is the point (−2, −2).


2 2

2 22 23 3

22 2 232 1 2 1 1 13 9 3 9 9 9 9

2 2 225 51 13 3 9 3

9 6 9 6 23

9 9 23

x x y y

x x y y

x x y y

x y


, and

radius 53

.




2 2

2 21 14 4

1 14 4

2 21 12 2

2 2 91 12 2 4

4 4 7

4 4 –

7 4 4

4 4 – 9

–

x x y y

x x y y

x y

x y


, and

radius 32

.

39. The equations of the three circles are 2 2( 7) ( 4) 25x y , 2 2( 9) ( 4) 169x y , and 2 2( 3) ( 9) 100x y . From the graph of the



2 2

2 2

2 2

( 7) ( 4) 25

(3 7) (1 4) 25

4 3 25 25 25

x y

2 2

2 2

2 2

( 9) ( 4) 169

(3 9) (1 4) 169

12 5 169 169 169

x y

2 2

2 2

2 2

( 3) ( 9) 100

(3 3) (1 9) 100

6 ( 8) 100 100 100

x y


40. The three equations are 2 2( 3) ( 1) 5x y , 2 2( 5) ( 4) 36x y , and 2 2( 1) ( 4) 40x y . From the graph of the



2 2

2 2

2 2

( 3) ( 1) 5

(5 3) (2 1) 5

2 1 5 5 5

x y

2 2

2 2

2

( 5) ( 4) 36

(5 5) (2 4) 36

6 36 36 36

x y

2 2

2 2

2 2

( 1) ( 4) 40

(5 1) (2 4) 40

6 ( 2) 40 40 40

x y


41. From the graph of the three circles, it appears that the epicenter is located at (−2, −2).


2 2

2 2

2 2

( 2) ( 1) 25

( 2 2) ( 2 1) 25

( 4) ( 3) 2525 25

x y

2 2

2 2

2 2

( 2) ( 2) 16

( 2 2) ( 2 2) 16

0 ( 4) 1616 16

x y

2 2

2 2

2 2

( 1) ( 2) 9

( 2 1) ( 2 2) 9

( 3) 0 99 9

x y

(−2, −2) satisfies all three equations, so the epicenter is at (−2, −2).



42. From the graph of the three circles, it appears that the epicenter is located at (5, 0).


2 2

2 2

2 2

( 2) ( 4) 25

(5 2) (0 4) 25

3 ( 4) 2525 25

x y

2 2

2 2

2 2

( 1) ( 3) 25

(5 1) (0 3) 25

4 3 2525 25

x y

2 2

2 2

2 2

( 3) ( 6) 100

(5 3) (0 6) 100

8 6 100100 100

x y


43. The radius of this circle is the distance from the center C(3, 2) to the x-axis. This distance is 2, so r = 2.

2 2 2

2 2

( – 3) ( – 2) 2

( – 3) ( – 2) 4

x yx y

44. The radius is the distance from the center C(–4, 3) to the point P(5, 8).

2 2

2 2

[5 – (– 4)] (8 – 3)

9 5 106

r


2 3

[ – (– 4)] ( – 3) ( 106)

( 4) ( – 3) 106

x yx y

45. Label the points P(x, y) and Q(1, 3).

If ( , ) 4d P Q , 2 21 3 4x y

2 21 3 16.x y

If x = y, then we can either substitute x for y or y for x. Substituting x for y we solve the following:

2 2

2 2

2

2

2

1 3 16

1 2 9 6 16

2 8 10 16

2 8 6 0

4 3 0

x xx x x x

x xx xx x

To solve this equation, we can use the quadratic formula with a = 1, b = –4, and c = –3.

24 4 4 1 3

2 1

4 16 12 4 28

2 2

4 2 72 7

2

x

Because x = y, the points are

2 7, 2 7 and 2 – 7, 2 7 .

46. Let P(–2, 3) be a point which is 8 units from Q(x, y). We have

2 2( , ) 2 3 8d P Q x y

2 22 3 64.x y

Because x + y = 0, x = –y. We can either substitute x for y or y for x. Substituting

x for y we solve the following:

22

2 2

2 2

2

2

2 3 64

2 3 64

4 4 9 6 64

2 10 13 64

2 10 51 0

x x

x xx x x x

x xx x

To solve this equation, use the quadratic formula with a = 2, b = 10, and c = –51.

210 10 4 2 51

2 2

10 100 408

410 4 12710 508

4 410 2 127 5 127

4 2

x

Because y x the points are

5 127 5 127,

2 2

and

–5 127 5 127, .

2 2



47. Let P(x, y) be a point whose distance from

A(1, 0) is 10 and whose distance from

B(5, 4) is 10 . d(P, A) = 10 , so

2 2(1 ) (0 ) 10x y 2 2(1 ) 10.x y ( , ) 10, sod P B

2 2(5 ) (4 ) 10x y 2 2(5 ) (4 ) 10.x y Thus,

2 2 2 2

2 2

2 2

(1 ) (5 ) (4 )

1 2

25 10 16 81 2 41 10 8

8 40 85

x y x yx x y

x x y yx x yy xy x

Substitute 5 – x for y in the equation 2 2(1 ) 10x y and solve for x. 2 2

2 2

2 2

2

(1 ) (5 ) 10

1 2 25 10 10

2 12 26 10 2 12 16 0

6 8 0 ( 2)( 4) 02 0 or 4 0

2 or 4

x xx x x x

x x x xx x x xx x

x x

To find the corresponding values of y use the equation y = 5 – x. If x = 2, then y = 5 – 2 = 3. If x = 4, then y = 5 – 4 = 1. The points satisfying the conditions are (2, 3) and (4, 1).

48. The circle of smallest radius that contains the points A(1, 4) and B(–3, 2) within or on its boundary will be the circle having points A and B as endpoints of a diameter. The center will be M, the midpoint:

1 3 4 2 2 6, ,

2 2 2 2

(−1, 3).

The radius will be the distance from M to either A or B:

2 2

2 2

( , ) [1 ( 1)] (4 3)

2 1 4 1 5

d M A


22 2

2 2

1 3 5

1 3 5.

x y

x y

49. Label the points A(3, y) and B(–2, 9). If d(A, B) = 12, then

2 2

2 2

2 2 2

2

2

2 3 9 12

5 9 12

5 9 12

25 81 18 144

18 38 0

y

y

yy y

y y

Solve this equation by using the quadratic formula with a = 1, b = –18, and c = –38:

218 18 4 1 38

2 1

18 324 152 18 476

2 1 2

y

18 4 119 18 2 119

9 1192 2

The values of y are 9 119 and 9 119 .

50. Because the center is in the third quadrant, the

radius is 2 , and the circle is tangent to both

axes, the center must be at ( 2, 2). Using the center-radius of the equation of a circle, we have

2 2 2

2 2

2 2 2

2 2 2.

x y

x y

51. Let P(x, y) be the point on the circle whose distance from the origin is the shortest. Complete the square on x and y separately to write the equation in center-radius form:

2 2

2 2

2 2

16 14 88 0

16 64 14 4988 64 49

( 8) ( 7) 25

x x y yx x y y

x y

So, the center is (8, 7) and the radius is 5.

2 2( , ) 8 7 113d C O . Because the

length of the radius is 5, ( , ) 113 5d P O .



52. Using compasses, draw circles centered at Wickenburg, Kingman, Phoenix, and Las Vegas with scaled radii of 50, 75, 105, and 180 miles respectively. The four circles should intersect at the location of Nothing.


3 –9–1+5 4 6 , , (2, – 3).

2 2 2 2

54. Use points C(2, –3) and P(–1, 3).

22

2 2

( , ) –1 – 2 3 – –3

–3 6 9 36

45 3 5

d C P

The radius is 3 5.

55. Use points C(2, –3) and Q(5, –9).

22

22

( , ) 5 – 2 –9 – –3

3 –6 9 36

45 3 5

d C Q

The radius is 3 5.

56. Use the points P(–1, 3) and Q(5, –9).

Because 2 2( , ) 5 – –1 –9 – 3d P Q

226 –12 36 144 180

6 5 ,the radius is 1

( , ).2

d P Q Thus

16 5 3 5.

2r

57. The center-radius form for this circle is 2 2 2( – 2) ( 3) (3 5)x y 2 2( – 2) ( 3) 45.x y

58. Label the endpoints of the diameter P(3, –5) and Q(–7, 3). The midpoint M of the segment joining P and Q has coordinates

3 (–7) –5 3 4 –2, , (–2, –1).

2 2 2 2

The center is C(–2, –1). To find the radius, we can use points C(–2, –1) and P(3, –5)

2 2

22

( , ) 3 – –2 –5 – –1

5 –4 25 16 41

d C P

We could also use points C(–2, –1).and Q(–7, 3).

2 2

2 2

( , ) 7 – –2 3 – –1

5 4 25 16 41

d C Q

We could also use points P(3, –5) and Q(–7, 3) to find the length of the diameter. The length of the radius is one-half the length of the diameter.

22

2 2

( , ) 7 – 3 3 – –5

10 8 100 64

164 2 41

d P Q

1 1( , ) 2 41 41

2 2d P Q


2 2 2

2 2

[ – (–2)] [ – (–1)] ( 41)

( 2) ( 1) 41

x yx y

59. Label the endpoints of the diameter P(–1, 2) and Q(11, 7). The midpoint M of the segment joining P and Q has coordinates

1 11 2 7 9, 5, .

2 2 2

The center is 92

5, .C To find the radius, we

can use points 92

5,C and P(−1, 2).

22 92

22 5 169 132 4 2

( , ) 5 – –1 2

6

d C P

We could also use points 92

5,C and

Q(11, 7).

22 92

22 5 169 132 4 2

( , ) 5 11 – 7

6

d C Q




(continued)


2 2

2 2

, 1 11 2 7

12 5

169 13

d P Q

1 1 13, 13

2 2 2d P Q


2 22 9 132 2

22 9 1692 4

5

5

x y

x y

60. Label the endpoints of the diameter P(5, 4) and Q(−3, −2). The midpoint M of the segment joining P and Q has coordinates

5 ( 3) 4 ( 2), 1, 1 .

2 2

The center is C(1, 1). To find the radius, we can use points C(1, 1) and P(5, 4).

2 2

2 2

( , ) 5 1 4 1

4 3 25 5

d C P

We could also use points C(1, 1) and Q(−3, −2).

2 2

2 2

( , ) 1 3 1 –2

4 3 25 5

d C Q


2 2

2 2

, 5 3 4 2

8 6 100 10

d P Q

1 1( , ) 10 5

2 2d P Q


2 2 2

2 2

1 1 5

1 1 25

x yx y

61. Label the endpoints of the diameter P(1, 4) and Q(5, 1). The midpoint M of the segment joining P and Q has coordinates

52

1 5 4 1, 3, .

2 2

The center is 52

3, .C


2 2 2 21 5 4 1 4 3 25 5.


5 .


2 22 5 52 2

22 5 252 4

3

3

x y

x y

62. Label the endpoints of the diameter P(−3, 10) and Q(5, −5). The midpoint M of the segment joining P and Q has coordinates

52

3 5 10 ( 5), 1, .

2 2

The center is 52

1, .C


22 2 23 5 10 5 8 15

289 17.


17 .


2 22 5 172 2

22 5 2892 4

1

1

x y

x y

Section 2.3 Functions

1. The domain of the relation

3,5 , 4,9 , 10,13 is 3, 4,10 .

2. The range of the relation in Exercise 1 is

5,9,13 .

3. The equation y = 4x – 6 defines a function with independent variable x and dependent variable y.

4. The function in Exercise 3 includes the ordered pair (6, 18).

5. For the function 4 2,f x x

2 4 2 2 8 2 10.f

6. For the function ,g x x 9 9 3.g


domain 0, .




range 0, .

For exercises 9 and 10, use this graph.

9. The largest open interval over which the function graphed here increases is , 3 .

10. The largest open interval over which the function graphed here decreases is 3, .



13. Two ordered pairs, namely (2, 4) and (2, 6), have the same x-value paired with different y-values, so the relation is not a function.

14. Two ordered pairs, namely (9, −2) and (9, 1), have the same x-value paired with different y-values, so the relation is not a function.





19. Two sets of ordered pairs, namely (1, 1) and (1, −1) as well as (2, 4) and (2, −4), have the same x-value paired with different y-values, so the relation is not a function. domain: {0, 1, 2}; range: {−4, −1, 0, 1, 4}

20. The relation is not a function because the x-value 3 corresponds to two y-values, 7 and 9. This correspondence can be shown as follows.

domain: {2, 3, 5}; range: {5, 7, 9, 11}

21. The relation is a function because for each different x-value there is exactly one y-value. domain: {2, 3, 5, 11, 17}; range: {1, 7, 20}

22. The relation is a function because for each different x-value there is exactly one y-value. domain: {1, 2, 3, 5}; range: {10, 15, 19, 27}


Domain: {0, −1, −2}; range: {0, 1, 2}




Domain: {0, 1, 2}; range: {0, −1, −2}

25. The relation is a function because for each different year, there is exactly one number for visitors. domain: {2010, 2011, 2012, 2013} range: {64.9, 63.0, 65.1, 63.5}

26. The relation is a function because for each basketball season, there is only one number for attendance. domain: {2011, 2012, 2013, 2014} range: {11,159,999, 11,210,832, 11,339,285, 11,181,735}


28. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: , ; range: , 4

29. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y. domain: 3, ; range: ,

30. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of x corresponds to two values of y. domain: [−4, 4]; range: [−3, 3]


32. This graph represents a function. If you pass a vertical line through the graph, one x-value corresponds to only one y-value. domain: [−2, 2]; range: [0, 4]

33. 2y x represents a function because y is always found by squaring x. Thus, each value of x corresponds to just one value of y. x can be any real number. Because the square of any real number is not negative, the range would be zero or greater.

domain: , ; range: 0,

34. 3y x represents a function because y is always found by cubing x. Thus, each value of x corresponds to just one value of y. x can be any real number. Because the cube of any real number could be negative, positive, or zero, the range would be any real number.



satisfy 6.x y This equation does not represent a function. Because x is equal to the sixth power of y, the values of x are nonnegative. Any real number can be raised to the sixth power, so the range of the relation is all real numbers.

domain: 0, range: ,




satisfy 4.x y This equation does not represent a function. Because x is equal to the fourth power of y, the values of x are nonnegative. Any real number can be raised to the fourth power, so the range of the relation is all real numbers.

domain: 0, range: ,

37. 2 5y x represents a function because y is found by multiplying x by 2 and subtracting 5. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers. Because y is twice x, less 5, y also may be any real number, and so the range is also all real numbers.


38. 6 4y x represents a function because y is found by multiplying x by −6 and adding 4. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers. Because y is −6 times x, plus 4, y also may be any real number, and so the range is also all real numbers.


39. By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into x + y < 3 corresponds to many values of y. The ordered pairs (1, –2), (1, 1), (1, 0), (1, −1), and so on, all satisfy the inequality. Note that the points on the graphed line do not satisfy the inequality and only indicate the boundary of the solution set. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers.


40. By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into x − y < 4 corresponds to many values of y. The ordered pairs (1, −1), (1, 0), (1, 1), (1, 2), and so on, all satisfy the inequality. Note that the points on the graphed line do not satisfy the inequality and only indicate the boundary of the solution set. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers.




41. For any choice of x in the domain of ,y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have 0.x Because the radical is nonnegative, the range is also zero or greater.

domain: 0, ; range: 0,


,y x there is exactly one corresponding value of y, so this equation defines a function. Because the quantity under the square root cannot be negative, we have 0.x The outcome of the radical is nonnegative, when you change the sign (by multiplying by −1), the range becomes nonpositive. Thus the range is zero or less.

domain: 0, ; range: , 0

43. Because 2xy can be rewritten as 2 ,xy

we can see that y can be found by dividing x into 2. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 0. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.

domain: , 0 0, ;

range: , 0 0,

44. Because xy = −6 can be rewritten as 6 ,xy

we can see that y can be found by dividing x into −6. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 0. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.

domain: , 0 0, ;

range: , 0 0,



4 1 0 4 1x x 14

.x Because the

radical is nonnegative, the range is also zero or greater.

domain: 1

4, ; range: 0,





7 2 0 2 7x x 7 72 2

or .x x

Because the radical is nonnegative, the range is also zero or greater.

domain: 72

, ; range: 0,


,xy we

find y by subtracting 3, then dividing into 2. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 3. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.

domain: ,3 3, ;

range: , 0 0,


,xy we

find y by subtracting 5, then dividing into −7. This process produces one value of y for each value of x in the domain, so this equation is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = 5. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers except zero.

domain: ,5 5, ;

range: , 0 0,

49. B. The notation f(3) means the value of the dependent variable when the independent variable is 3.

50. Answers will vary. An example is: The cost of gasoline depends on the number of gallons used; so cost is a function of number of gallons.

51.

3 4

0 3 0 4 0 4 4

f x xf

52.

3 4

3 3 3 4 9 4 13

f x xf

53.

2

2

4 1

2 2 4 2 1

4 8 1 11

g x x xg

54.

2

2

4 1

10 10 4 10 1100 40 1 59

g x x xg

55. 1 1

3 3

3 4

3 4 1 4 3

f x xf

56. 7 7

3 3

3 4

3 4 7 4 11

f x xf

57.

2

21 1 12 2 2

1 114 4

4 1

4 1

2 1

g x x x

g

58.

2

21 1 14 4 4

1 116 16

4 1

4 1

1 1

g x x x

g

59.

3 4

3 4

f x xf p p



60.

2

2

4 1

4 1

g x x xg k k k

61.

3 4

3 4 3 4

f x xf x x x

62.

2

2

2

4 1

4 1

4 1

g x x xg x x x

x x

63.

3 4

2 3 2 43 6 4 3 2

f x xf x x

x x

64.

3 4

4 3 4 43 12 4 3 8

f x xf a a

a a

65.

3 4

2 3 3 2 3 46 9 4 6 13

f x xf m m

m m

66.

3 4

3 2 3 3 2 49 6 4 9 10

f x xf t t

t t

67. (a) 2 2f (b) 1 3f

68. (a) 2 5f (b) 1 11f

69. (a) 2 15f (b) 1 10f

70. (a) 2 1f (b) 1 7f

71. (a) 2 3f (b) 1 3f

72. (a) 2 3f (b) 1 2f

73. (a) 2 0f (b) 0 4f

(c) 1 2f (d) 4 4f

74. (a) 2 5f (b) 0 0f

(c) 1 2f (d) 4 4f

75. (a) 2 3f (b) 0 2f

(c) 1 0f (d) 4 2f

76. (a) 2 3f (b) 0 3f

(c) 1 3f (d) 4 3f

77. (a)

1 13 3

3 123 12

12

34 4

x yy x

xy

y x f x x

(b) 13

3 3 4 1 4 3f

78. (a)

1 14 4

4 88 4

8 48

42 2

x yx y

x yx y

y x f x x

(b) 3 3 8 514 4 4 4 4

3 3 2 2f

79. (a)

2

2

2

2 3

2 3

2 3

y x xy x x

f x x x

(b) 23 2 3 3 32 9 3 3 18

f

80. (a)

2

2

2

3 2

3 2

3 2

y x xy x x

f x x x

(b) 23 3 3 3 23 9 3 2 32

f

81. (a)

8 84 43 3 3 3

4 3 84 3 8

4 8 34 8

3

x yx y

x yx y

y x f x x

(b) 8 84 12 43 3 3 3 3

3 3f

82. (a)

9 92 25 5 5 5

2 5 95 2 9

2 9

5

x yy x

xy

y x f x x

(b) 9 6 9 1525 5 5 5 5

3 3 3f

83. 3 4f



84. Because 20.2 0.2 3 0.2 1f

= 0.04 + 0.6 + 1 = 1.64, the height of the rectangle is 1.64 units. The base measures 0.3 − 0.2 = 0.1 unit. Because the area of a rectangle is base times height, the area of this rectangle is 0.1(1.64) = 0.164 square unit.


which is −4.


which is −3.

87. (a) 2, 0 (b) , 2

(c) 0,

88. (a) 3, 1 (b) 1,

(c) , 3

89. (a) , 2 ; 2,

(b) 2, 2 (c) none

90. (a) 3,3 (b) , 3 ; 3,

(c) none

91. (a) 1, 0 ; 1,

(b) , 1 ; 0, 1

(c) none

92. (a) , 2 ; 0, 2

(b) 2, 0 ; 2,

(c) none

93. (a) Yes, it is the graph of a function.

(b) [0, 24]

(c) When t = 8, y = 1200 from the graph. At 8 A.M., approximately 1200 megawatts is being used.

(d) The most electricity was used at 17 hr or 5 P.M. The least electricity was used at 4 A.M.

(e) 12 1900f

At 12 noon, electricity use is about 1900 megawatts.

(f) increasing from 4 A.M. to 5 P.M.; decreasing from midnight to 4 A.M. and from 5 P.M. to midnight

94. (a) At t = 2, y = 240 from the graph. Therefore, at 2 seconds, the ball is 240 feet high.

(b) At y = 192, x = 1 and x = 5 from the graph. Therefore, the height will be 192 feet at 1 second and at 5 seconds.

(c) The ball is going up from 0 to 3 seconds and down from 3 to 7 seconds.

(d) The coordinate of the highest point is (3, 256). Therefore, it reaches a maximum height of 256 feet at 3 seconds.

(e) At x = 7, y = 0. Therefore, at 7 seconds, the ball hits the ground.

95. (a) At t = 12 and t = 20, y = 55 from the graph. Therefore, after about 12 noon until about 8 P.M. the temperature was over 55º.

(b) At t = 6 and t = 22, y = 40 from the graph. Therefore, until about 6 A.M. and after 10 P.M. the temperature was below 40º.

(c) The temperature at noon in Bratenahl, Ohio was 55º. Because the temperature in Greenville is 7º higher, we are looking for the time at which Bratenahl, Ohio was at or above 48º. This occurred at approximately 10 A.M and 8:30 P.M.

(d) The temperature is just below 40° from midnight to 6 A.M., when it begins to rise until it reaches a maximum of just below 65° at 4 P.M. It then begins to fall util it reaches just under 40° again at midnight.

96. (a) At t = 8, y = 24 from the graph. Therefore, there are 24 units of the drug in the bloodstream at 8 hours.

(b) The level increases between 0 and 2 hours after the drug is taken and decreases between 2 and 12 hours after the drug is taken.

(c) The coordinates of the highest point are (2, 64). Therefore, at 2 hours, the level of the drug in the bloodstream reaches its greatest value of 64 units.

(d) After the peak, y = 16 at t = 10. 10 hours – 2 hours = 8 hours after the peak. 8 additional hours are required for the level to drop to 16 units.



(e) When the drug is administered, the level is 0 units. The level begins to rise quickly for 2 hours until it reaches a maximum of 64 units. The level then begins to decrease gradually until it reaches a level of 12 units, 12 hours after it was administered.

Section 2.4 Linear Functions

1. B; ( )f x = 3x + 6 is a linear function with y-intercept (0, 6).

2. H; x = 9 is a vertical line.

3. C; ( )f x = −8 is a constant function.

4. G; 2x – y = –4 or y = 2x + 4 is a linear equation with x-intercept (–2, 0) and y-intercept (0, 4).

5. A; ( )f x = 5x is a linear function whose graph passes through the origin, (0, 0). f(0) = 5(0) = 0.

6. D; 2( )f x x is a function that is not linear.

7. –3m matches graph C because the line falls rapidly as x increases.

8. m = 0 matches graph A because horizontal lines have slopes of 0.

9. m = 3 matches graph D because the line rises rapidly as x increases.

10. m is undefined for graph B because vertical lines have undefined slopes.


(0) 0 – 4 – 4 :f y-intercept

0 – 4 4 :x x x-intercept Graph the line through (0, –4) and (4, 0).



(0) –0 4 4 :f y-intercept

0 – 4 4 :x x x-intercept Graph the line through (0, 4) and (4, 0).


13. 12

( ) – 6f x x

Use the intercepts.

12

(0) 0 – 6 – 6 :f y-intercept

1 12 2

0 – 6 6 12 :x x x x-intercept

Graph the line through (0, –6) and (12, 0).


14. 23

( ) 2f x x

Use the intercepts.

23


2 23 3

0 2 –2 3 : -interceptx x x x

Graph the line through (0, 2) and (–3, 0).




15. ( ) 3f x x The x-intercept and the y-intercept are both zero. This gives us only one point, (0, 0). If x = 1, y = 3(1) = 3 . Another point is (1, 3). Graph the line through (0, 0) and (1, 3).


16. –2f x x

The x-intercept and the y-intercept are both zero. This gives us only one point, (0, 0). If x = 3, y = –2(3) = –6, so another point is (3, –6). Graph the line through (0, 0) and (3, –6).


17. ( ) – 4f x is a constant function.

The graph of ( ) 4f x is a horizontal line with a y-intercept of –4.

domain: , ; range: {–4}

18. ( ) 3f x is a constant function whose graph is a horizontal line with y-intercept of 3.


19. ( ) 0f x is a constant function whose graph is the x-axis.


20. 9f x x


21. 4 3 12x y Use the intercepts.

4 0 3 12 3 124 : -intercept

y yy y

4 3 0 12 4 123 : -intercept

x xx x

Graph the line through (0, 4) and (−3, 0).


22. 2 5 10;x y Use the intercepts.

2 0 5 10 5 102 : -intercept

y yy y

2 5 0 10 2 105 : -intercept

x xx x





23. 3 4 0y x Use the intercepts.

3 4 0 0 3 0 0 : -intercepty y y y

3 0 4 0 4 0 0 : -interceptx x x x


3 4 3 0 3 12 03 12 4

y yy y



24. 3 2 0x y Use the intercepts. 3 0 2 0 2 0 0 : -intercepty y y y

3 2 0 0 3 0 0 :x x x x-intercept


3 2 2 0 6 2 02 6 3

y yy y

Graph the line through (0, 0) and (2, −3):


25. x = 3 is a vertical line, intersecting the x-axis at (3, 0).


26. x = –4 is a vertical line intersecting the x-axis at (–4, 0).


27. 2x + 4 = 0 2 4 2x x is a vertical line intersecting the x-axis at (–2, 0).


28. 3 6 0 –3 6 2x x x is a vertical

line intersecting the x-axis at (2, 0).


29. 5 0 5x x is a vertical line intersecting the x-axis at (5, 0).




30. 3 0 3x x is a vertical line

intersecting the x-axis at 3,0 .


31. y = 5 is a horizontal line with y-intercept 5. Choice A resembles this.

32. y = –5 is a horizontal line with y-intercept –5. Choice C resembles this.

33. x = 5 is a vertical line with x-intercept 5. Choice D resembles this.

34. x = –5 is a vertical line with x-intercept –5. Choice B resembles this.

35.

36.

37.

38.

39. The rise is 2.5 feet while the run is 10 feet so

the slope is 2.5 110 4

0.25 25% . So A =

0.25, 2.510

C , D = 25%, and 14

E are all

expressions of the slope.

40. The pitch or slope is 14

. If the rise is 4 feet

then 1 rise 4

4 run x or x = 16 feet. So 16 feet in

the horizontal direction corresponds to a rise of 4 feet.

41. Through (2, –1) and (–3, –3) Let 1 1 22, –1, –3, and x y x 2 –3.y

Then rise –3 – (–1) –2y and

run –3 – 2 –5.x


.run –5 5

ymx

42. Through (−3, 4) and (2, −8) Let 1 1 2 23, 4, 2, and –8.x y x y

Then rise –8 – 4 –12y and

run 2 – ( 3) 5.x


.run 5 5

ymx

43. Through (5, 8) and (3, 12) Let 1 1 2 25, 8, 3, and 12.x y x y

Then rise 12 8 4y and

run 3 5 2.x

The slope is rise 4

2.run 2

ymx

44. Through (5, –3) and (1, –7)

1 1 2 2Let 5, –3, 1, and –7.x y x y

Then rise –7 – (–3) – 4y and

run 1 – 5 – 4.x

The slope is 4

1.4

ymx

45. Through (5, 9) and (–2, 9)

2 1

2 1

9 – 9 00

–2 5 –7

y yymx x x

46. Through (–2, 4) and (6, 4)

2 1

2 1

4 4 00

6 (–2) 8

y yymx x x

47. Horizontal, through (5, 1) The slope of every horizontal line is zero, so m = 0.

48. Horizontal, through (3 , 5) The slope of every horizontal line is zero, so m = 0.

49. Vertical, through (4, –7) The slope of every vertical line is undefined; m is undefined.



50. Vertical, through (–8, 5) The slope of every vertical line is undefined; m is undefined.

51. (a) 3 5y x Find two ordered pairs that are solutions to the equation. If x = 0, then 3 0 5 5.y y

If x = −1, then 3 1 5 3 5 2.y y y

Thus two ordered pairs are (0, 5) and (−1, 2)

2 1

2 1

rise 2 5 33.

run 1 0 1

y ymx x

(b)

52. 2 4y x Find two ordered pairs that are solutions to the equation. If 0,x then 2 0 4y

4.y If 1,x then 2 1 4y

2 4 2.y y Thus two ordered pairs

are 0, 4 and 1, 2 .

2 1

2 1

2 4rise 22.

run 1 0 1

y ymx x

(b)


If 3,y then 2 3 3 6 3x x

2.x Thus two ordered pairs are 0,0 and

2, 3 .

2 1

2 1

rise 3 0 3.

run 2 0 2

y ymx x

(b)


If 4,x then 4 5 4 4 20y y

5.y Thus two ordered pairs are 0,0 and

4, 5 .

2 1

2 1

rise 5 0 5.

run 4 0 4

y ymx x

(b)


5.y If 0,y then 5 2 0 10x

5 10x 2.x

Thus two ordered pairs are 0, 5 and 2,0 .

2 1

2 1

0 5rise 5.

run 2 0 2

y ymx x

(b)




3 12y 4.y If 0,y then

4 3 0 12 4 12x x 3.x Thus two

ordered pairs are 0, 4 and 3,0 .

2 1

2 1

rise 0 4 4.

run 3 0 3

y ymx x

(b)

57. Through (–1, 3), 32

m

First locate the point (–1, 3). Because the

slope is 32

, a change of 2 units horizontally (2

units to the right) produces a change of 3 units vertically (3 units up). This gives a second point, (1, 6), which can be used to complete the graph.

58. Through (–2, 8), 25

m . Because the slope is

25

, a change of 5 units horizontally (to the

right) produces a change of 2 units vertically (2 units up). This gives a second point (3, 10), which can be used to complete the graph. Alternatively, a change of 5 units to the left produces a change of 2 units down. This gives the point (−7, 6).

59. Through (3, –4), 13

–m . First locate the point

(3, –4). Because the slope is 13

– , a change of

3 units horizontally (3 units to the right) produces a change of –1 unit vertically (1 unit down). This gives a second point, (6, –5), which can be used to complete the graph.

60. Through (–2, –3), 34

–m . Because the slope

is 3 –34 4

– , a change of 4 units horizontally

(4 units to the right) produces a change of –3 units vertically (3 units down). This gives a second point (2, –6), which can be used to complete the graph.

61. Through 12

, 4 , m = 0.


12

, 4 .


62. Through 32

, 2 , m = 0.


32

, 2 .



63. Through 52

, 3 , undefined slope. The slope

is undefined, so the line is vertical,

intersecting the x-axis at 52

, 0 .

64. Through 94

, 2 , undefined slope. The slope is

undefined, so the line is vertical, intersecting

the x-axis at 94

, 0 .


mb a

20 4 16$4

0 4 4

(thousand) per year. The

value of the machine is decreasing $4000 each year during these years.


mb a

200 0 200$50

4 0 4

per month. The amount

saved is increasing $50 each month during these months.

67. The graph is a horizontal line, so the average rate of change (slope) is 0. The percent of pay raise is not changing—it is 3% each year.

68. The graph is a horizontal line, so the average rate of change (slope) is 0. That means that the number of named hurricanes remained the same, 10, for the four consecutive years shown.

69. 2562 5085 2523

2012 1980 3278.8 thousand per year

f b f am

b a

The number of high school dropouts decreased by an average of 78.8 thousand per year from 1980 to 2012.

70. 1709 5302

2013 20063593

$513.297

f b f am

b a

Sales of plasma flat-panel TVs decreased by an average of $513.29 million per year from 2006 to 2013.

71. (a) The slope of –0.0167 indicates that the average rate of change of the winning time for the 5000 m run is 0.0167 min less. It is negative because the times are generally decreasing as time progresses.

(b) The Olympics were not held during World Wars I (1914−1919) and II (1939–1945).

(c) 0.0167 2000 46.45 13.05 miny

The model predicts a winning time of 13.05 minutes. The times differ by 13.35 − 13.05 = 0.30 min.

72. (a) From the equation, the slope is 200.02. This means that the number of radio stations increased by an average of 200.02 per year.

(b) The year 2018 is represented by x = 68. 200.02 68 2727.7 16,329.06y

According to the model, there will be about 16,329 radio stations in 2018.

73. 2013 2008 335,652 270,334

2013 2008 2013 200865,318

13,063.65

f f

The average annual rate of change from 2008 through 2013 is about 13,064 thousand.

74. 2014 2006 3.74 4.53

2014 2006 2014 20060.79

0.0998

f f

The average annual rate of change from 2006 through 2014 is about –0.099.



75. (a) 56.3 138

2013 200381.7

8.1710

f b f am

b a

The average rate of change was −8.17 thousand mobile homes per year.

(b) The negative slope means that the number of mobile homes decreased by an average of 8.17 thousand each year from 2003 to 2013.

76. 2013 1991 26.6 61.8

2013 1991 2013 199135.2

1.622

f f

There was an average decrease of 1.6 births per thousand per year from 1991 through 2013.

77. (a) 10 500C x x

(b) 35R x x

(c) 35 10 500

35 10 500 25 500

P x R x C xx xx x x

(d) 10 500 35

500 2520

C x R xx x

xx

20 units; do not produce

78. (a) 150 2700C x x

(b) 280R x x

(c) 280 150 2700

280 150 2700130 2700

P x R x C xx xx xx

(d) 150 2700 280

2700 13020.77 or 21 units

C x R xx x

xx

21 units; produce

79. (a) 400 1650C x x

(b) 305R x x

(c) 305 400 1650

305 400 165095 1650

P x R x C xx xx xx

(d) 400 1650 305

95 1650 095 1650

17.37 units

C x R xx xx

xx

This result indicates a negative “break-even point,” but the number of units produced must be a positive number. A calculator graph of the lines

1 400 1650y C x x and

2 305y R x x in the window

[0, 70] × [0, 20000] or solving the inequality 305 400 1650x x will show

that R x C x for all positive values

of x (in fact whenever x is greater than −17.4). Do not produce the product because it is impossible to make a profit.

80. (a) 11 180C x x

(b) 20R x x

(c) 20 11 180

20 11 180 9 180

P x R x C xx xx x x

(d) 11 180 20

180 920

C x R xx x

xx

20 units; produce

81. 200 1000 2401000 40 25C x R x x x

x x

The break-even point is 25 units. (25) 200 25 1000 $6000C which is the

same as (25) 240 25 $6000R

C x

R x

Chapter 2 Quiz (Sections 2.1–2.4) 203


82. 220 1000 2401000 20 50C x R x x x

x x

The break-even point is 50 units instead of 25 units. The manager is not better off because twice as many units must be sold before beginning to show a profit.

83. The first two points are A(0, –6) and B(1, –3). 3 – (–6) 3

31 – 0 1

m

84. The second and third points are B(1, –3) and C(2, 0).

0 – (–3) 33

2 – 1 1m

85. If we use any two points on a line to find its slope, we find that the slope is the same in all cases.

86. The first two points are A(0, –6) and B(1, –3). 2 2

2 2

( , ) (1 – 0) [–3 – (– 6)]

1 3 1 9 10

d A B

87. The second and fourth points are B(1, –3) and D(3, 3).

2 2

2 2

( , ) (3 – 1) [3 – (–3)]

2 6 4 36

40 2 10

d B D

88. The first and fourth points are A(0, –6) and D(3, 3).

2 2

2 2

( , ) (3 – 0) [3 – (– 6)]

3 9 9 81

90 3 10

d A D

89. 10 2 10 3 10; The sum is 3 10, which is equal to the answer in Exercise 88.

90. If points A, B, and C lie on a line in that order, then the distance between A and B added to the distance between B and C is equal to the distance between A and C.

91. The midpoint of the segment joining A(0, –6) and G(6, 12) has coordinates

0 6 6 122 2

, 6 62 2

, 3,3 . The midpoint is

M(3, 3), which is the same as the middle entry in the table.

92. The midpoint of the segment joining E(4, 6) and F(5, 9) has coordinates

4 5 6 9 9 152 2 2 2

, , = (4.5, 7.5). If the

x-value 4.5 were in the table, the corresponding y-value would be 7.5.


1.

2 22 1 2 1

2 2

2 2

( , )

8 ( 4) 3 2

( 4) ( 5) 16 25 41

d A B x x y y

2. To find an estimate for 2006, find the midpoint of (2004, 6.55) and (2008, 6.97:

2004 2008 6.55 6.97,

2 22006, 6.76

M

The estimated enrollment for 2006 was 6.76 million. To find an estimate for 2010, find the midpoint of (2008, 6.97) and (2012, 7.50):

2008 2012 6.97 7.50,

2 22010, 7.235

M

The estimated enrollment for 2006 was about 7.24 million.

3.

4.


2 2

2 2

( 4 4) ( 8 16) 3 4 16

( 2) ( 4) 17

x x y yx y

The radius is 17 and the midpoint of the circle is (2, −4).

6. From the graph, f(−1) is 2.

7. Domain: ( , ); range: [0, )



8. (a) The largest open interval over which f is decreasing is ( , 3) .

(b) The largest open interval over which f is increasing is 3, .

(c) There is no interval over which the function is constant.

9. (a) 11 5 6 3

5 1 4 2m

(b) 4 4 0

01 ( 7) 6

m

(c) 4 12 16

6 6 0m

the slope is

undefined.

10. The points to use are (2009, 10,602) and (2013, 15,884). The average rate of change is

15,884 10,602 5282

1320.52013 2009 4

The average rate of change was 1320.5 thousand cars per year. This means that the number of new motor vehicles sold in the United States increased by an average of 1320.5 thousand per year from 2009 to 2013.


1. The graph of the line 3 4 8y x has

slope 4 and passes through the point (8, 3).

2. The graph of the line 2 7y x has slope –2 and y-intercept (0, 7).

3. The vertical line through the point (–4, 8) has equation x = –4.

4. The horizontal line through the point (–4, 8) has equation y = 8.

For exercises 5 and 6, 67

6 7 0 7 6x y y x y x

5. Any line parallel to the graph of 6 7 0x y

must have slope 67.

6. Any line perpendicular to the graph of

6 7 0x y must have slope 76.

7. 14

2y x is graphed in D.

The slope is 14

and the y-intercept is (0, 2).

8. 4x + 3y = 12 or 3y = –4x + 12 43

or – 4y x

is graphed in B. The slope is 43

– and the

y-intercept is (0, 4).

9. 32

– 1 1y x is graphed in C. The slope

is 32

and a point on the graph is (1, –1).

10. y = 4 is graphed in A. y = 4 is a horizontal line with y-intercept (0, 4).

11. Through (1, 3), m = –2. Write the equation in point-slope form.

1 1– – 3 –2 –1y y m x x y x

Then, change to standard form. 3 –2 2 2 5y x x y

12. Through (2, 4), m = –1 Write the equation in point-slope form.

1 1– – 4 –1 – 2y y m x x y x

Then, change to standard form. – 4 – 2 6y x x y

13. Through (–5, 4), 32

m


32

– 4 5y x

Change to standard form. 2 – 4 –3 52 – 8 –3 – 15

3 2 –7

y xy x

x y

14. Through 34

(– 4, 3), m


34

– 3 4y x

Change to standard form. 4 3 3 4

4 12 3 123 4 24 or 3 4 24

y xy xx y x y

15. Through (–8, 4), undefined slope Because undefined slope indicates a vertical line, the equation will have the form x = a. The equation of the line is x = –8.

16. Through (5, 1), undefined slope This is a vertical line through (5, 1), so the equation is x = 5.

17. Through (5, −8), m = 0 This is a horizontal line through (5, −8), so the equation is y = −8.

18. Through (−3, 12), m = 0 This is a horizontal line through (−3, 12), so the equation is y = 12.



19. Through (–1, 3) and (3, 4) First find m.

4 – 3 1

3 – (–1) 4m


14

– 4 – 34 16 3

4 134 13

y xy xx yx y

20. Through (2, 3) and (−1, 2) First find m.

2 3 1 1

1 2 3 3m


13

3 – 23 – 9 2

3 73 7

y xy x

x yx y

21. x-intercept (3, 0), y-intercept (0, –2) The line passes through (3, 0) and (0, –2). Use these points to find m.

–2 – 0 2

0 – 3 3m


– 2.y x

22. x-intercept (–4, 0), y-intercept (0, 3) The line passes through the points (–4, 0) and (0, 3). Use these points to find m.

3 – 0 3

0 – (–4) 4m


3.y x

23. Vertical, through (–6, 4) The equation of a vertical line has an equation of the form x = a. Because the line passes through (–6, 4), the equation is x = –6. (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)

24. Vertical, through (2, 7) The equation of a vertical line has an equation of the form x = a. Because the line passes through (2, 7), the equation is x = 2. (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)

25. Horizontal, through (−7, 4) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (−7, 4), the equation is y = 4.

26. Horizontal, through (−8, −2) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (−8, −2), the equation is y = −2.

27. m = 5, b = 15 Using slope-intercept form, we have

5 15.y x

28. m = −2, b = 12 Using slope-intercept form, we have

2 12.y x

29. Through (−2, 5), slope = −4

5 4 2

5 4 25 4 8

4 3

y xy xy x

y x

30. Through (4, −7), slope = −2 7 2 4

7 2 82 1

y xy x

y x


0,

These represent 32

0 and .m b


3 32 2

0 .y x y


0,

These represent 54

0 and .m b


5 54 4

0 .y x y

33. The line x + 2 = 0 has x-intercept (–2, 0). It does not have a y-intercept. The slope of his line is undefined. The line 4y = 2 has

y-intercept 12

0, . It does not have an

x-intercept. The slope of this line is 0.

34. (a) The graph of y = 3x + 2 has a positive slope and a positive y-intercept. These conditions match graph D.



(b) The graph of y = –3x + 2 has a negative slope and a positive y-intercept. These conditions match graph B.

(c) The graph of y = 3x – 2 has a positive slope and a negative y-intercept. These conditions match graph A.

(d) The graph of y = –3x – 2 has a negative slope and a negative y-intercept. These conditions match graph C.

35. y = 3x – 1 This equation is in the slope-intercept form, y = mx + b. slope: 3; y-intercept: (0, –1)

36. y = –2x + 7 slope: –2; y-intercept: (0, 7)

37. 4x – y = 7 Solve for y to write the equation in slope-intercept form. – – 4 7 4 – 7y x y x slope: 4; y-intercept: (0, –7)

38. 2x + 3y = 16 Solve the equation for y to write the equation in slope-intercept form.

1623 3

3 –2 16 –y x y x

slope: 23

– ; y-intercept: 163

0,

39. 3 34 4

4 –3 – or 0y x y x y x

slope: 34

; y-intercept (0, 0)

40. 1 12 2

2 or 0y x y x y x

slope is 12



12

2 – 4 – 2y x y x

slope: 12

– ; y-intercept: (0, −2)




13

3 – 9 – 3y x y x

slope: 13

– ; y-intercept: (0,−3)

43. 32

1 0y x

Solve the equation for y to write the equation in slope-intercept form.

3 32 2

1 0 1y x y x

slope: 32


44. (a) Use the first two points in the table, A(–2, –11) and B(–1, –8).

–8 – (–11) 33

–1 – (–2) 1m

(b) When x = 0, y = −5. The y-intercept is (0, –5).

(c) Substitute 3 for m and –5 for b in the slope-intercept form.

3 – 5y mx b y x

45. (a) The line falls 2 units each time the x value increases by 1 unit. Therefore the slope is −2. The graph intersects the y-axis at the point (0, 1) and intersects the

x-axis at 12



, 0 .


46. (a) The line rises 2 units each time the x value increases by 1 unit. Therefore the slope is 2. The graph intersects the y-axis at the point (0, −1) and intersects the

x-axis at 12



, 0 .


47. (a) The line falls 1 unit each time the x value increases by 3 units. Therefore the slope

is 13

. The graph intersects the y-axis at

the point (0, 2), so the y-intercept is (0, 2). The graph passes through (3, 1) and will fall 1 unit when the x value increases by 3, so the x-intercept is (6, 0).


2.y x

48. (a) The line rises 3 units each time the x value increases by 4 units. Therefore the

slope is 34

. The graph intersects the

y-axis at the point (0, −3) and intersects the x-axis at (4, 0), so the y-intercept is (0, −3) and the x-intercept is 4.


3.y x

49. (a) The line falls 200 units each time the x value increases by 1 unit. Therefore the slope is −200. The graph intersects the y-axis at the point (0, 300) and intersects

the x-axis at 32



, 0 .


50. (a) The line rises 100 units each time the x value increases by 5 units. Therefore the slope is 20. The graph intersects the y-axis at the point (0, −50) and intersects

the x-axis at 52



, 0 .




51. (a) through (–1, 4), parallel to x + 3y = 5 Find the slope of the line x + 3y = 5 by writing this equation in slope-intercept form.

513 3

3 5 3 – 5–

x y y xy x

The slope is 13

.

Because the lines are parallel, 13

is also

the slope of the line whose equation is to

be found. Substitute 13

m , 1 –1,x

and 1 4y into the point-slope form.

1 11313

– –

4 – –1

– 4 13 –12 – –1 3 11

y y m x xy xy x

y x x y


3 – 11 –y x y x

52. (a) through (3, –2), parallel to 2x – y = 5 Find the slope of the line 2x – y = 5 by writing this equation in slope-intercept form. 2 – 5 – –2 5

2 – 5x y y x

y x

The slope is 2. Because the lines are parallel, the slope of the line whose equation is to be found is also 2. Substitute m = 2, 1 3x , and 1 2y into the point-slope form.

1 1 – –

2 2 – 3 2 2 – 6–2 –8 or 2 – 8

y y m x xy x y x

x y x y

(b) Solve for y. y = 2x – 8

53. (a) through (1, 6), perpendicular to 3x + 5y = 1 Find the slope of the line 3x + 5y = 1 by writing this equation in slope-intercept form.

3 15 5

3 5 1 5 –3 1–

x y y xy x


. The slope of


,

because 3 55 3

1. Substitute 53

m ,

1 1,x and 1 6y into the point-slope form.

53

– 6 ( –1)3( – 6) 5( –1)3 – 18 5 – 5

–13 5 – 3 or 5 – 3 –13

y xy xy x

x y x y


3 5 13y x y x

54. (a) through (–2, 0), perpendicular to 8x – 3y = 7 Find the slope of the line 8x – 3y = 7 by writing the equation in slope-intercept form.

8 73 3

8 – 3 7 –3 –8 7–

x y y xy x


. The slope of


,

because 8 33 8

1.

Substitute 38

m , 1 2,x and 1 0y

into the point-slope form. 38

– 0 – ( 2)8 –3( 2)8 –3 – 6 3 8 –6

y xy xy x x y


3 38 4

8 –3 – 6 –

–

y x y xy x

55. (a) through (4, 1), parallel to y = −5 Because y = −5 is a horizontal line, any line parallel to this line will be horizontal and have an equation of the form y = b. Because the line passes through (4, 1), the equation is y = 1.


56. (a) through 2, 2 , parallel to y = 3.

Because y = 3 is a horizontal line, any line parallel to this line will be horizontal and have an equation of the form y = b. Because the line passes through 2, 2 ,

the equation is y = –2.

(b) The slope-intercept form is y = −2

57. (a) through (–5, 6), perpendicular to x = –2. Because x = –2 is a vertical line, any line perpendicular to this line will be horizontal and have an equation of the form y = b. Because the line passes through (–5, 6), the equation is y = 6.




58. (a) Through (4, –4), perpendicular to x = 4 Because x = 4 is a vertical line, any line perpendicular to this line will be horizontal and have an equation of the form y = b. Because the line passes through (4, –4), the equation is y = –4.

(b) The slope-intercept form is y = –4.

59. (a) Find the slope of the line 3y + 2x = 6.

23

3 2 6 3 –2 6– 2

y x y xy x

Thus, 23

– .m A line parallel to

3y + 2x = 6 also has slope 23

– .

Solve for k using the slope formula.

12

2 1 2–

4 33 2

–4 3

3 23 4 3 4 –

4 39 2 49 2 8

2 1

k

k

k kk

kk

k k

(b) Find the slope of the line 2y – 5x = 1.

5 12 2

2 5 1 2 5 1y x y xy x

Thus, 52

.m A line perpendicular to 2y

– 5x = 1 will have slope 25

– , because

5 22 5

– –1.

Solve this equation for k.

72

3 2

4 53 2

5 4 5 44 515 2 415 2 82 7

k

k kk

kk

k k

60. (a) Find the slope of the line 2 – 3 4.x y

2 43 3

2 – 3 4 –3 –2 4–

x y y xy x

Thus, 23

.m A line parallel to

2x – 3y = 4 also has slope 23

. Solve for r

using the slope formula. – 6 2 – 6 2

– 4 – 2 3 – 6 3

– 6 26 6

– 6 36 4 2

r r

r

r r

(b) Find the slope of the line x + 2y = 1.

1 12 2

2 1 2 – 1–

x y y xy x

Thus, 12

.m A line perpendicular to

the line x + 2y = 1 has slope 2, because 12

– (2) –1. Solve for r using the slope

formula. – 6 – 6

2 2– 4 – 2 – 6

– 6 –12 – 6

r r

r r


7703 6312 1391463.67

3 0 3m


463.67 6312.y x


463.67 4 6312 8166.68y

The model predicts that average tuition and fees were $8166.68 in 2013. This is $96.68 more than the actual amount.


7136 6312 824412

2 0 2m

The y-intercept is (0, 6312), so the equation of the line is 412 6312.y x


412 4 6312 7960y

The model predicts that average tuition and fees were $7960 in 2013. This is $110 less than the actual amount.




24525 22036 2489622.25

4 0 4m


622.25 22,036.y x

The slope of the line indicates that the

average tuition increase is about $622 per year from 2009 through 2013.

(b) The year 2012 corresponds to x = 3. 622.25 3 22,036 23,902.75y

According to the model, average tuition and fees were $23,903 in 2012. This is $443 more than the actual amount $23,460.

(c) Using the linear regression feature, the equation of the line of best fit is

653 21,634.y x

64. (a) See the graph in the answer to part (b).There appears to be a linear relationship between the data. The farther the galaxy is from Earth, the faster it is receding.

(b) Using the points (520, 40,000) and (0, 0), we obtain

40,000 – 0 40,00076.9.

520 – 0 520m

The equation of the line through these two points is y = 76.9x.

(c) 76.9 60,00060,000

78076.9

x

x x

According to the model, the galaxy Hydra is approximately 780 megaparsecs away.

(d) 11

1110 9

9.5 10

9.5 101.235 10 12.35 10

76.9

Am

A

Using m = 76.9, we estimate that the age of the universe is approximately 12.35 billion years.

(e) 11

10 9

119

9.5 101.9 10 or 19 10

509.5 10

9.5 10100

A

A

The range for the age of the universe is between 9.5 billion and 19 billion years.

65. (a) The ordered pairs are (0, 32) and (100, 212).

The slope is 212 – 32 180 9

.100 – 0 100 5

m

Use 91 1 5

( , ) (0,32) and x y m in the

point-slope form.

1 195959 95 5

– ( – )

– 32 ( – 0)

– 32

32 32

y y m x xy xy x

y x F C

(b)

95

59

32

5 9 325 9 160 9 5 –1609 5( – 32) ( – 32)

F CF CF C C FC F C F

(c) 59

( – 32)9 5( – 32) 9 5 – 1604 –160 – 40

F C F FF F F FF F

F = C when F is –40º.



66. (a) The ordered pairs are (0, 1) and (100, 3.92). The slope is

3.92 – 1 2.920.0292

100 – 0 100m and 1.b

Using slope-intercept form we have 0.0292 1y x or ( ) 0.0292 1.p x x

(b) Let x = 60. P(60) = 0.0292(60) + 1 = 2.752 The pressure at 60 feet is approximately 2.75 atmospheres.

67. (a) Because we want to find C as a function of I, use the points (12026, 10089) and (14167, 11484), where the first component represents the independent variable, I. First find the slope of the line.

11484 10089 13950.6516

14167 12026 2141m

Now use either point, say (12026, 10089), and the point-slope form to find the equation.

–10089 0.6516( –12026)–10089 0.6516 – 7836

0.6516 2253

C IC I

C I

(b) Because the slope is 0.6516, the marginal propensity to consume is 0.6516.

68. D is the only possible answer, because the x-intercept occurs when y = 0. We can see from the graph that the value of the x-intercept exceeds 10.

69. Write the equation as an equivalent equation with 0 on one side: 2 7 4 2x x x 2 7 4 2 0x x x . Now graph

2 7 4 2y x x x in the window [–5, 5] × [–5, 5] to find the x-intercept:

Solution set: {3}

70. Write the equation as an equivalent equation with 0 on one side: 7 2 4 5 3 1x x x 7 2 4 5 3 1 0x x x . Now graph

7 2 4 5 3 1y x x x in the window [–5, 5] × [–5, 5] to find the x-intercept:

Solution set: {1}

71. Write the equation as an equivalent equation with 0 on one side: 3 2 1 2 2 5x x

3 2 1 2 2 5 0x x . Now graph

3 2 1 2 2 5y x x in the window

[–5, 5] × [–5, 5] to find the x-intercept:

Solution set: 12

or 0.5

72. Write the equation as an equivalent equation with 0 on one side:

4 3 4 2 2 3 6 2x x x x

4 3 4 2 2 3 6 2 0x x x x .

Now graph 4 3 4 2 2 3 6 2y x x x x in the

window [–2, 8] × [–5, 5] to find the x-intercept:

Solution set: {4}

73. (a) 2 5 22 10 2

10 212

x xx x

xx

Solution set: {12}



(b) Answers will vary. Sample answer: The solution does not appear in the standard viewing window x-interval [10, –10]. The minimum and maximum values must include 12.

74. Rewrite the equation as an equivalent equation with 0 on one side.

3 2 6 4 8 2

6 18 4 8 2 0

x x xx x x

Now graph y = −6x − 18 − (−4x + 8 −2x) in the window [–10, 10] × [–30, 10].

The graph is a horizontal line that does not

intersect the x-axis. Therefore, the solution set is . We can verify this algebraically.

3 2 6 4 8 26 18 6 8 0 26

x x xx x

Because this is a false statement, the solution set is .

75. A(–1, 4), B(–2, –1), C(1, 14)

For A and B, 1 4 5

52 ( 1) 1

m

For B and C, 14 ( 1) 15

51 ( 2) 3

m


51 ( 1) 2

m


76. A(0, −7), B(–3, 5), C(2, −15)

For A and B, 5 ( 7) 12

43 0 3

m

For B and C, 15 5 20

42 ( 3) 5

m

For A and C, 15 ( 7) 8

42 0 2

m


77. A(−1, −3), B(−5, 12), C(1, −11)

For A and B, 12 ( 3) 15

5 ( 1) 4m

For B and C, 11 12 23

1 ( 5) 6m

For A and C, 11 ( 3) 8

41 ( 1) 2

m

Since all three slopes are not the same, the points are not collinear.

78. A(0, 9), B(–3, –7), C(2, 19)

For A and B, 7 9 16 16

3 0 3 3m

For B and C, 19 ( 7) 26

2 ( 3) 5m


52 0 2

m

Because all three slopes are not the same, the points are not collinear.

79. 2 21 1 1

2 2 21 1 1

( , ) ( – 0) ( – 0)d O P x m x

x m x

80. 2 22 2 2

2 2 22 2 2

( , ) ( – 0) ( – 0)d O Q x m x

x m x

81. 2 22 1 2 2 1 1( , ) ( – ) ( – )d P Q x x m x m x

82.

2 2 2

2 22 2 2 2 2 2

1 1 1 2 2 2

22 2

2 1 2 2 1 1

2 2 2 2 2 21 1 1 2 2 2

2 22 1 2 2 1 1

[ ( , )] [ ( , )] [ ( , )]

– –

– –

d O P d O Q d P Q

x m x x m x

x x m x m x

x m x x m x

x x m x m x

2 2 2 2 2 21 1 1 2 2 2

2 2 2 22 2 1 1 2 2

2 21 2 1 2 1 1

2 1 1 2 1 2

1 2 1 2 1

2

20 2 2

2 2 0

x m x x m xx x x x m x

m m x x m xx x m m x x

m m x x x x

83. 1 2 1 2 1 2

1 2 1 2

–2 – 2 0–2 ( 1) 0m m x x x x

x x m m

84. 1 2 1 2–2 ( 1) 0x x m m

Because 1 20 and 0, x x we have

1 2 1 21 0 implying that –1.m m m m



85. If two nonvertical lines are perpendicular, then the product of the slopes of these lines is –1.

Summary Exercises on Graphs, Circles, Functions, and Equations

1. (3, 5), (2, 3)P Q

(a)

2 2

2 2

( , ) (2 3) ( 3 5)

1 8

1 64 65

d P Q


5 33 2 5 2 5, , ,1

2 2 2 2 2

.


82 – 3 1

m

Use either point and the point-slope form. – 5 8 – 3y x

Change to slope-intercept form. 5 8 24 8 19y x y x

2. P(–1, 0), Q(4, –2)

(a) 2 2

2 2

( , ) [4 – (–1)] (–2 – 0)

5 (–2)

25 4 29

d P Q


–1 4 0 (–2) 3 2, ,

2 2 2 23

, 1 .2

(c) First find m: 2 – 0 2 2

4 – 1 5 5m


25

– 0 – 1y x


2 25 5

5 2 15 2 2

y xy xy x

3. P(–2, 2), Q(3, 2)

(a) 22

2 2

( , ) [3 – (– 2)] 2 2

5 0 25 0 25 5

d P Q


– 2 3 2 2,

2 2

1 4 1, , 2 .

2 2 2


03 – 2 5

m

All lines that have a slope of 0 are horizontal lines. The equation of a horizontal line has an equation of the form y = b. Because the line passes through (3, 2), the equation is y = 2.

4. 2 2, 2 ,P 2,3 2Q

(a)

2 2

2 2

( , ) 2 – 2 2 3 2 – 2

– 2 2 2

2 8 10

d P Q


2 2 2 2 3 2,

2 2

3 2 4 2 3 2, , 2 2 .

2 2 2


22 2 2 2

m


– 2 2 – 2 2y x


– 2 2 4 2 2 5 2y x y x

5. P(5, –1), Q(5, 1)

(a) 2 2

2 2

( , ) (5 – 5) [1 – (–1)]

0 2 0 4 4 2

d P Q


5 5 –1 1,

2 2

10 0

, 5,0 .2 2



(c) First find m. 1 – 1 2

undefined5 – 5 0

m

All lines that have an undefined slope are vertical lines. The equation of a vertical line has an equation of the form x = a. The line passes through (5, 1), so the equation is x = 5 . (Because the slope of a vertical line is undefined, this equation cannot be written in slope-intercept form.)

6. (1, 1), ( 3, 3)P Q

(a)

2 2

2 2

( , ) ( 3 – 1) ( 3 –1)

4 4

16 16 32 4 2

d P Q


1 3 1 3,

2 2

2 2,

2 2

1, 1 .

(c) First find m: 3 –1 4

13 –1 4

m

Use either point and the point-slope form. –1 1 –1y x

Change to slope-intercept form. 1 1y x y x

7. 2 3,3 5 , 6 3,3 5P Q

(a)

2 2

2 2

( , ) 6 3 – 2 3 3 5 – 3 5

4 3 0 48 4 3

d P Q


2 3 6 3 3 5 3 5,

2 2

8 3 6 5, 4 3, 3 5 .

2 2


06 3 2 3 4 3

m

All lines that have a slope of 0 are horizontal lines. The equation of a horizontal line has an equation of the form y = b. Because the line passes

through 2 3,3 5 , the equation is

3 5.y

8. P(0, –4), Q(3, 1)

(a) 2 2

2 2

( , ) (3 – 0) [1 – (– 4)]

3 5 9 25 34

d P Q


0 3 – 4 1 3 3 3 3, , , .

2 2 2 2 2 2


3 0 3m


– 4.y x

9. Through 2,1 and 4, 1

First find m: 1 –1 2 1

4 – (–2) 6 3m


13

– 1 – 4y x


1 13 3

3 1 4 3 3 4

3 1

y x y xy x y x



10. the horizontal line through (2, 3) The equation of a horizontal line has an equation of the form y = b. Because the line passes through (2, 3), the equation is y = 3.

11. the circle with center (2, –1) and radius 3

22 2

2 2

( – 2) – (–1) 3

( – 2) ( 1) 9

x yx y

12. the circle with center (0, 2) and tangent to the x-axis The distance from the center of the circle to the x-axis is 2, so r = 2.

2 2 2 2 2( – 0) ( – 2) 2 ( – 2) 4x y x y

13. the line through (3, 5) with slope 56


56

– 5 3 y x

Change to standard form.

5 156 6

5 56 2

6 5 5 3 6 30 5 15

6 5 15

y x y xy x y x

y x

14. the vertical line through (−4, 3) The equation of a vertical line has an equation of the form x = a. Because the line passes through (−4, 3), the equation is x = −4.

15. a line through (–3, 2) and parallel to the line 2x + 3y = 6 First, find the slope of the line 2x + 3y = 6 by writing this equation in slope-intercept form.

23

2 3 6 3 2 6 2x y y x y x

The slope is 23

. Because the lines are

parallel, 23

is also the slope of the line

whose equation is to be found. Substitute 23

m , 1 3x , and 1 2y into the point-

slope form.

21 1 3

23

– – 2 3

3 2 2 3 3 6 2 – 6

3 –2

y y m x x y xy x y x

y x y x

16. a line through the origin and perpendicular to the line 3 4 2x y

First, find the slope of the line 3 4 2x y by writing this equation in slope-intercept form.

3 32 14 4 4 2

3 4 2 4 –3 2x y y xy x y x


. The slope of any

line perpendicular to this line is 43

, because

343 4

1. Using slope-intercept form we

have 4 43 3

0 or .y x y x




(continued)


2 2

2 2

2 2

4 2 4

4 4 2 1 4 4 1

2 1 9

x x y y

x x y y

x y

Yes, it is a circle. The circle has its center at (2, −1) and radius 3.

18. 2 26 10 36 0x x y y Complete the square on x and y separately.

2 2

2 2

2 2

6 10 36

6 9 10 25 –36 9 25

3 5 2

x x y y

x x y y

x y



2 2

2 2

2 2

12 20

12 36 –20 36

6 16

x x y

x x y

x y



2 2

2 2

2 2

2 16 61

2 1 16 64 –61 1 64

1 8 4

x x y y

x x y y

x y

Yes, it is a circle. The circle has its center at (−1, −8) and radius 2.


2 2

2 2

2 2

2 10

2 1 –10 1

1 9

x x y

x x y

x y


22. 2 2 – 8 9 0x y y Complete the square on x and y separately.

2 2

2 2

22

– 8 9

– 8 16 9 16

– 4 25

x y y

x y y

x y


23. The equation of the circle is 2 2 2( 4) ( 5) 4 .x y

Let y = 2 and solve for x: 2 2 2( 4) (2 5) 4x 2 2 2 2( 4) ( 3) 4 ( 4) 7x x

4 7 4 7x x

The points of intersection are 4 7, 2 and

4 7, 2

24. Write the equation in center-radius form by completing the square on x and y separately:

2 2

2 2

2 2

2 2

10 24 144 0

10 24 144 0

( 10 25) ( 24 144) 25

( 5) ( 12) 25

x y x yx x y y

x x y yx y

The center of the circle is (5, 12) and the radius is 5. Now use the distance formula to find the distance from the center (5, 12) to the origin:

2 22 1 2 1

2 2(5 0) (12 0) 25 144 13

d x x y y

The radius is 5, so the shortest distance from the origin to the graph of the circle is 13 − 5 = 8.




(continued)


4 4 4 24 6 .y x y x y x

x can be any real number, so the domain is all real numbers and the range is also all real numbers. domain: , ; range: ,

(b) Each value of x corresponds to just one value of y. 4 6x y represents a function.

3 31 14 2 4 2

y x f x x

3 31 1 24 2 2 2 2

2 2 1f

26. (a) The equation can be rewritten as 2 5 .y x y can be any real number.

Because the square of any real number is

not negative, 2y is never negative. Taking the constant term into consideration, domain would be 5, .

domain: 5, ; range: ,


the relation, 2 5y x does not represent a function.

27. (a) 2 22 25x y is a circle centered at

(−2, 0) with a radius of 5. The domain will start 5 units to the left of –2 and end 5 units to the right of –2. The domain will be [−2 − 5, 2 + 5] = [−7, 3]. The range will start 5 units below 0 and end 5 units above 0. The range will be [0 − 5, 0 + 5] = [−5, 5].


the relation, 2 22 25x y does not

represent a function.


2 22 3 .y x y x x can be

any real number. Because the square of

any real number is not negative, 212

x is

never negative. Taking the constant term into consideration, range would be

32

, .

domain: , ; range: 32

,

(b) Each value of x corresponds to just one

value of y. 2 2 3x y represents a function.

2 23 31 12 2 2 2

y x f x x

2 3 3 31 1 4 12 2 2 2 2 2 2

2 2 4f

Section 2.6 Graphs of Basic Functions

1. The equation 2f x x matches graph E.

The domain is , .

2. The equation of f x x matches graph G.

The function is increasing on 0, .

3. The equation 3f x x matches graph A.

The range is , .

4. Graph C is not the graph of a function.

Its equation is 2x y .

5. Graph F is the graph of the identity function. Its equation is .f x x

6. The equation f x x matches graph B.

1.5 1f

7. The equation 3f x x matches graph H.

No, there is no interval over which the function is decreasing.

8. The equation of f x x matches graph D.

The domain is 0, .

9. The graph in B is discontinuous at many points. Assuming the graph continues, the range would be {…, −3, −2, −1, 0, 1, 2, 3, …}.



10. The graphs in E and G decrease over part of the domain and increase over part of the domain. They both increase over 0, and

decrease over , 0 .



13. The function is continuous over the interval

0, .

14. The function is continuous over the interval , 0 .

15. The function has a point of discontinuity at (3, 1). It is continuous over the interval ,3 and the interval 3, .

16. The function has a point of discontinuity at x = 1. It is continuous over the interval ,1

and the interval 1, .

17. 2 if –1( )

–1 if –1x xf x x x

(a) (–5) 2(–5) –10f

(b) (–1) 2(–1) –2f

(c) f(0) = 0 –1 = –1

(d) f(3) = 3 – 1 = 2

18. – 2 if 3( )

5 – if 3x xf x x x

(a) f(–5) = –5 – 2 = –7

(b) f(–1) = –1 – 2 = –3

(c) f(0) = 0 – 2 = –2

(d) f(3) = 5 – 3 = 2

19. 2 if – 4

( ) – if – 4 23 if 2

x xf x x x

x x

(a) f(–5) = 2 + (–5) = –3

(b) f(–1) = – (–1) = 1

(c) f(0) = –0 = 0

(d) (3) 3 3 9f

20. –2 if –33 –1 if – 3 2– 4 if 2

x xf x x x

x x

(a) f(–5) = –2(–5) = 10

(b) f(–1) = 3(–1) – 1 = –3 – 1 = –4

(c) f(0) = 3(0) – 1 = 0 – 1 = –1

(d) f(3) = –4(3) = –12

21. – 1 if 3( )

2 if 3x xf x x

Draw the graph of y = x – 1 to the left of x = 3, including the endpoint at x = 3. Draw the graph of y = 2 to the right of x = 3, and note that the endpoint at x = 3 coincides with the endpoint of the other ray.

22. 6 – if 3( )

3 if 3x xf x x

Graph the line y = 6 – x to the left of x = 3, including the endpoint. Draw y = 3 to the right of x = 3. Note that the endpoint at x = 3 coincides with the endpoint of the other ray.

23. 4 – if 2( )

1 2 if 2x xf x x x

Draw the graph of y = 4 − x to the left of x = 2, but do not include the endpoint. Draw the graph of y = 1 + 2x to the right of x = 2, including the endpoint.



24. 2 1 if 0( )

if 0x xf x x x

Graph the line y = 2x + 1 to the right of x = 0, including the endpoint. Draw y = x to the left of x = 0, but do not include the endpoint.

25. 3 if 1( )

1 if 1xf x x

Graph the line y = −3 to the left of x = 1, including the endpoint. Draw y = −1 to the right of x = 1, but do not include the endpoint.

26. 2 if 1( )

2 if 1xf x x

Graph the line y = – 2 to the left of x = 1, including the endpoint. Draw y = 2 to the right of x = 1, but do not include the endpoint.


27. 2 if – 4

( ) – if – 4 53 if 5

x xf x x x

x x

Draw the graph of y = 2 + x to the left of –4, but do not include the endpoint at x = 4. Draw the graph of y = –x between –4 and 5, including both endpoints. Draw the graph of y = 3x to the right of 5, but do not include the endpoint at x = 5.

28. –2 if –3

( ) 3 –1 if – 3 2– 4 if 2

x xf x x x

x x

Graph the line y = –2x to the left of x = –3, but do not include the endpoint. Draw y = 3x – 1 between x = –3 and x = 2, and include both endpoints. Draw y = –4x to the right of x = 2, but do not include the endpoint. Notice that the endpoints of the pieces do not coincide.

29. 21

212

– +2 if 2( )

if 2

x xf x

x x

Graph the curve 212

2y x to the left of

x = 2, including the endpoint at (2, 0). Graph

the line 12

y x to the right of x = 2, but do

not include the endpoint at (2, 1). Notice that the endpoints of the pieces do not coincide.

30. 3

25 if 0

( )if 0

x xf xx x

Graph the curve 3 5y x to the left of x = 0, including the endpoint at (0, 5). Graph

the line 2y x to the right of x = 0, but do not include the endpoint at (0, 0). Notice that the endpoints of the pieces do not coincide.



31. 2

2 if 5 1( ) 2 if 1 0

2 if 0 2

x xf x x

x x

Graph the line 2y x between x = −5 and x = −1, including the left endpoint at (−5, −10), but not including the right endpoint at (−1, −2). Graph the line y = −2 between x = −1 and x = 0, including the left endpoint at (−1, −2) and not including the right endpoint at (0, −2). Note that (−1, −2) coincides with the first two sections, so it is included. Graph

the curve 2 2y x from x = 0 to x = 2, including the endpoints at (0, −2) and (2, 2).

Note that (0, −2) coincides with the second two sections, so it is included. The graph ends at x = −5 and x = 2.

32.

2

2

0.5 if 4 2( ) if 2 2

4 if 2 4

x xf x x x

x x

Graph the curve 20.5y x between x = −4 and x = −2, including the endpoints at (−4, 8) and (−2, 2). Graph the line y x between x = −2 and x = 2, but do not include the endpoints at (−2, −2) and (2, 2). Graph the

curve 2 4y x from x = 2 to x = 4, including the endpoints at (2, 0) and (4, 12). The graph ends at x = −4 and x = 4.

33.

3

2

3 if 2 0( ) 3 if 0 1

4 if 1 3

x xf x x x

x x x

Graph the curve 3 3y x between x = −2 and x = 0, including the endpoints at (−2, −5) and (0, 3). Graph the line y = x + 3 between x = 0 and x = 1, but do not include the endpoints at (0, 3) and (1, 4). Graph the curve

24y x x from x = 1 to x = 3, including the endpoints at (1, 4) and (3, −2). The graph ends at x = −2 and x = 3.

34. 2

312

2 if 3 1

( ) 1 if 1 2

1 if 2 3

x xf x x x

x x

Graph the curve y = −2x to from x = −3 to x = −1, including the endpoint (−3, 6), but not including the endpoint (−1, 2). Graph the

curve 2 1y x from x = −1 to x = 2, including the endpoints (−1, 2) and (2, 5).

Graph the curve 312

1y x from x = 2 to

x = 3, including the endpoint (3, 14.5) but not including the endpoint (2, 5). Because the endpoints that are not included coincide with endpoints that are included, we use closed dots on the graph.



35. The solid circle on the graph shows that the endpoint (0, –1) is part of the graph, while the open circle shows that the endpoint (0, 1) is not part of the graph. The graph is made up of parts of two horizontal lines. The function which fits this graph is

–1 if 0( )

1 if 0.xf x x

domain: , ; range: {–1, 1}

36. We see that y = 1 for every value of x except x = 0, and that when x = 0, y = 0. We can write the function as

1 if 0( )

0 if 0.xf x x

domain: , ; range: {0, 1}

37. The graph is made up of parts of two horizontal lines. The solid circle shows that the endpoint (0, 2) of the one on the left belongs to the graph, while the open circle shows that the endpoint (0, –1) of the one on the right does not belong to the graph. The function that fits this graph is

2 if 0( )

–1 if 1.xf x x

domain: ( , 0] (1, ); range: {–1, 2}

38. We see that y = 1 when 1x and that y = –1 when x > 2. We can write the function as

1 if –1( )

–1 if 2.xf x x

domain: (– , –1] (2, ); range: {–1, 1}

39. For 0x , that piece of the graph goes through the points (−1, −1) and (0, 0). The slope is 1, so the equation of this piece is y = x. For x > 0, that piece of the graph is a horizontal line passing through (2, 2), so its equation is y = 2. We can write the function as

if 0( ) .

2 if 0x xf x x

domain: (– , ) range: ( , 0] {2}

40. For x < 0, that piece of the graph is a horizontal line passing though (−3, −3), so the equation of this piece is y = −3. For 0x , the curve passes through (1, 1) and (4, 2), so the

equation of this piece is y x . We can

write the function as 3 if 0

( ) . if 0

xf x

x x

domain: (– , ) range: { 3} [0, )

41. For x < 1, that piece of the graph is a curve passes through (−8, −2), (−1, −1) and (1, 1), so

the equation of this piece is 3y x . The right piece of the graph passes through (1, 2) and

(2, 3). 2 3

11 2

m

, and the equation of the

line is 2 1 1y x y x . We can write

the function as 3 if 1( )

1 if 1x xf x

x x

domain: (– , ) range: ( ,1) [2, )

42. For all values except x = 2, the graph is a line. It passes through (0, −3) and (1, −1). The slope is 2, so the equation is y = 2x −3. At x = 2, the graph is the point (2, 3). We can write

the function as 3 if 2( ) .

2 3 if 2xf x x x

domain: (– , ) range: ( ,1) (1, )

43. f(x) = x

Plot points.

x –x f(x) = x

–2 2 2

–1.5 1.5 1

–1 1 1

–0.5 0.5 0

0 0 0

0.5 –0.5 –1

1 –1 –1

1.5 –1.5 –2

2 –2 –2

More generally, to get y = 0, we need 0 – 1 0 1 1 0.x x x To get y = 1, we need 1 – 2x

1 2 –2 1.x x Follow this pattern to graph the step function.

domain: , ; range: {...,–2,–1,0,1,2,...}



44. f(x) = x

Plot points.

x x f(x) = x

–2 −2 2

–1.5 −2 2

–1 −1 1

–0.5 −1 1

0 0 0

0.5 0 0

1 1 −1

1.5 1 –1

2 2 –2


domain: , ; range: {...,–2,–1,0,1,2,...}

45. f(x) = 2x


0 2 1 0 .x x


1 2 2 1.x x


2 2 3 1 .x x


domain: , ; range: {...,–2,–1,0,1,2,...}

46. g(x) = 2 1x


0 2 1 1 1 2 2 1.x x x


1 2 – 1 2 2 2 3 1 .x x x


domain: , ; range: {..., 2,–1,0,1,2,...}

47. The cost of mailing a letter that weighs more than 1 ounce and less than 2 ounces is the same as the cost of a 2-ounce letter, and the cost of mailing a letter that weighs more than 2 ounces and less than 3 ounces is the same as the cost of a 3-ounce letter, etc.

48. The cost is the same for all cars parking

between 12

hour and 1-hour, between 1 hour

and 12

1 hours, etc.

49.



50.

51. (a) For 0 8,x 49.8 34.2

1.958 0

m

,

so 1.95 34.2.y x For 8 13x ,

52.2 49.80.48

13 8m

, so the equation

is 52.2 0.48( 13)y x 0.48 45.96y x

(b) 1.95 34.2 if 0 8( )

0.48 45.96 if 8 13x xf x x x

52. When 0 3x , the slope is 5, which means that the inlet pipe is open, and the outlet pipe is closed. When 3 5x , the slope is 2, which means that both pipes are open. When 5 8x , the slope is 0, which means that both pipes are closed. When 8 10x , the slope is −3, which means that the inlet pipe is closed, and the outlet pipe is open.

53. (a) The initial amount is 50,000 gallons. The final amount is 30,000 gallons.

(b) The amount of water in the pool remained constant during the first and fourth days.

(c) (2) 45,000; (4) 40,000f f

(d) The slope of the segment between (1, 50000) and (3, 40000) is −5000, so the water was being drained at 5000 gallons per day.

54. (a) There were 20 gallons of gas in the tank at x = 3.

(b) The slope is steepest between t = 1 and t ≈ 2.9, so that is when the car burned gasoline at the fastest rate.

55. (a) There is no charge for additional length, so we use the greatest integer function. The cost is based on multiples of two

feet, so 2

( ) 0.8 xf x if 6 18x .

(b) 8.52

(8.5) 0.8 0.8(4) $3.20f

15.22

(15.2) 0.8 0.8(7) $5.60f

56. (a) 6.5 if 0 4

( ) 5.5 48 if 4 6–30 195 if 6 6.5

x xf x x x

x x

Draw a graph of y = 6.5x between 0 and 4, including the endpoints. Draw the graph of y = –5.5x + 48 between 4 and 6, including the endpoint at 6 but not the one at 4. Draw the graph of y = –30x + 195, including the endpoint at 6.5 but not the one at 6. Notice that the endpoints of the three pieces coincide.

(b) From the graph, observe that the snow depth, y, reaches its deepest level (26 in.) when x = 4, x = 4 represents 4 months after the beginning of October, which is the beginning of February.

(c) From the graph, the snow depth y is nonzero when x is between 0 and 6.5. Snow begins at the beginning of October and ends 6.5 months later, in the middle of April.

Section 2.7 Graphing Techniques


graph of 2y x down 3 units.


graph of 2y x up 5 units.


shifting the graph of 2y x to the left 4 units.


shifting the graph of 2y x to the right 7 units.


the graph of f x x across the x-axis.


the graph of f x x across the y-axis.




shift the graph of 3y x 2 units to the left and 3 units down.


shift the graph of 3y x 3 units to the right and 6 units up.

9. The graph of f x x is the same as the

graph of y x because reflecting it across

the y-axis yields the same ordered pairs.

10. The graph of 2x y is the same as the graph

of 2x y because reflecting it across the

x-axis yields the same ordered pairs.

11. (a) B; 2( 7)y x is a shift of 2 ,y x 7 units to the right.

(b) D; 2 7y x is a shift of 2 ,y x 7 units downward.

(c) E; 27y x is a vertical stretch of 2 ,y x by a factor of 7.

(d) A; 2( 7)y x is a shift of 2 ,y x 7 units to the left.

(e) C; 2 7y x is a shift of 2 ,y x 7 units upward.

12. (a) E; 34y x is a vertical stretch

of 3 ,y x by a factor of 4.

(b) C; 3y x is a reflection of 3 ,y x over the x-axis.

(c) D; 3y x is a reflection of 3 ,y x over the y-axis.

(d) A; 3 4y x is a shift of 3 ,y x 4 units to the right.

(e) B; 3 4y x is a shift of 3 ,y x 4 units down.

13. (a) B; 2 2y x is a shift of 2 ,y x 2 units upward.

(b) A; 2 2y x is a shift of 2 ,y x 2 units downward.

(c) G; 2( 2)y x is a shift of 2 ,y x 2 units to the left.

(d) C; 2( 2)y x is a shift of 2 ,y x 2 units to the right.

(e) F; 22y x is a vertical stretch of 2 ,y x by a factor of 2.

(f) D; 2y x is a reflection of 2 ,y x across the x-axis.

(g) H; 2( 2) 1y x is a shift of 2 ,y x 2 units to the right and 1 unit upward.

(h) E; 2( 2) 1y x is a shift of 2 ,y x 2 units to the left and 1 unit upward.

(i) I; 2( 2) 1y x is a shift of 2 ,y x 2 units to the left and 1 unit down.

14. (a) G; 3y x is a shift of ,y x 3 units to the left.

(b) D; 3y x is a shift of ,y x 3 units downward.

(c) E; 3y x is a shift of ,y x 3 units upward.

(d) B; 3y x is a vertical stretch of

,y x by a factor of 3.

(e) C; y x is a reflection of y x across the x-axis.

(f) A; 3y x is a shift of ,y x 3 units to the right.

(g) H; 3 2y x is a shift of ,y x 3 units to the right and 2 units upward.

(h) F; 3 2y x is a shift of ,y x 3 units to the left and 2 units upward.

(i) I; 3 2y x is a shift of ,y x 3 units to the right and 2 units downward.

15. (a) F; 2y x is a shift of y x 2 units

to the right.

(b) C; 2y x is a shift of y x 2 units

downward.

(c) H; 2y x is a shift of y x 2 units

upward.



(d) D; 2y x is a vertical stretch of y x

by a factor of 2.

(e) G; y x is a reflection of

y x across the x-axis.

(f) A; y x is a reflection of y x

across the y-axis.

(g) E; 2y x is a reflection of 2y x

across the x-axis. 2y x is a vertical

stretch of y x by a factor of 2.

(h) I; 2 2y x is a shift of y x 2

units to the right and 2 units upward.

(i) B; 2 2y x is a shift of y x 2

units to the left and 2 units downward.

16. The graph of 32 1 6f x x is the graph

of 3f x x stretched vertically by a factor

of 2, shifted left 1 unit and down 6 units.

17. 3f x x

x h x x 3f x x

−2 2 6

−1 1 3

0 0 0

1 1 3

2 2 6

18. 4f x x

x h x x 4f x x

−2 2 8

−1 1 4

0 0 0

1 1 4

2 2 8

19. 23

f x x

x h x x 23

f x x

−3 3 2

−2 2 43

−1 1 23

0 0 0

1 1 23

2 2 43

3 3 2

20. 34

f x x

x h x x 34

f x x

−4 4 3

−3 3 94

−2 2 32

−1 1 34

0 0 0

1 1 34

2 2 32

3 3 94

4 4 3




(continued)

21. 22f x x

x 2h x x 22f x x

−2 4 8

−1 1 2

0 0 0

1 1 2

2 4 8

22. 23f x x

x 2h x x 23f x x

−2 4 12

−1 1 3

0 0 0

1 1 3

2 4 12

23. 212

f x x

x 2h x x 212

f x x

−2 4 2

−1 1 12

0 0 0

1 1 12

2 4 2

24. 213

f x x

x 2h x x 213

f x x

−3 9 3

−2 4 43

−1 1 13

0 0 0

1 1 13

2 4 43

3 9 3



25. 212

f x x

x 2h x x 212

f x x

−3 9 92

−2 4 2

−1 1 12

0 0 0

1 1 12

2 4 2

3 9 92

26. 213

f x x

x 2h x x 213

f x x

−3 9 −3

−2 4 43

−1 1 13

0 0 0

1 1 13

2 4 43

3 9 −3

27. 3f x x

x h x x 3f x x

−2 2 −6

−1 1 −3

0 0 0

1 1 −3

2 2 −6

28. 2f x x

x h x x 2f x x

−2 2 −4

−1 1 −2

0 0 0

1 1 −2

2 2 −4

29. 12

h x x

x f x x 1

21 12 2

h x x

x x

−4 4 2

−3 3 32

−2 2 1

−1 1 12

0 0 0




(continued)

x f x x 1

21 12 2

h x x

x x

1 1 12

2 2 1

3 3 32

4 4 2

30. 13

h x x

x 13

f x x 1

31 13 3

h x x

x x

−3 3 1

−2 2 23

−1 1 13

0 0 0

1 1 13

2 2 23

3 3 1

31. 4h x x

x f x x 4 2h x x x

0 0 0

1 1 2

2 2 2 2

x f x x 4 2h x x x

3 3 2 3

4 2 4

32. 9h x x

x f x x 9 3h x x x

0 0 0

1 1 3

2 2 3 2

3 3 3 3

4 2 6

33. f x x

x h x x f x x

−4 2 −2

−3 3 3

−2 2 2

−1 1 −1

0 0 0



34. f x x

x h x x f x x

−3 3 −3

−2 2 −2

−1 1 −1

0 0 0

1 1 −1

2 2 −2

3 3 −3

35. (a) 4y f x is a horizontal translation

of f, 4 units to the left. The point that corresponds to (8, 12) on this translated function would be 8 4,12 4,12 .

(b) 4y f x is a vertical translation of f, 4 units up. The point that corresponds to (8, 12) on this translated function would be 8,12 4 8,16 .

36. (a) 14

y f x is a vertical shrinking of f, by

a factor of 14

. The point that corresponds

to (8, 12) on this translated function

would be 14

8, 12 8,3 .

(b) 4y f x is a vertical stretching of f, by

a factor of 4. The point that corresponds to (8, 12) on this translated function would be 8, 4 12 8, 48 .

37. (a) (4 )y f x is a horizontal shrinking of f, by a factor of 4. The point that corresponds to (8, 12) on this translated

function is 14

8 , 12 2, 12 .

(b) 14

y f x is a horizontal stretching of f, by a factor of 4. The point that corresponds to (8, 12) on this translated function is 8 4, 12 32, 12 .

38. (a) The point that corresponds to (8, 12) when reflected across the x-axis would be (8, −12).

(b) The point that corresponds to (8, 12) when reflected across the y-axis would be (−8, 12).

39. (a) The point that is symmetric to (5, –3) with respect to the x-axis is (5, 3).

(b) The point that is symmetric to (5, –3) with respect to the y-axis is (–5, –3).

(c) The point that is symmetric to (5, –3) with respect to the origin is (–5, 3).

40. (a) The point that is symmetric to (–6, 1) with respect to the x-axis is (–6, –1).

(b) The point that is symmetric to (–6, 1) with respect to the y-axis is (6, 1).

(c) The point that is symmetric to (–6, 1) with respect to the origin is (6, –1).

41. (a) The point that is symmetric to (–4, –2) with respect to the x-axis is (–4, 2).

(b) The point that is symmetric to (–4, –2) with respect to the y-axis is (4, –2).

(c) The point that is symmetric to (–4, –2) with respect to the origin is (4, 2).



42. (a) The point that is symmetric to (–8, 0) with respect to the x-axis is (–8, 0) because this point lies on the x-axis.

(b) The point that is symmetric to the point (–8, 0) with respect to the y-axis is (8, 0).

(c) The point that is symmetric to the point (–8, 0) with respect to the origin is (8, 0).

43. The graph of y = |x − 2| is symmetric with respect to the line x = 2.

44. The graph of y = −|x + 1| is symmetric with respect to the line x = −1.


2 2( ) 5 5y x x . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain

2 2 2( ) 2 2 2.y x y x y x The result is not the same as the original

equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.


4 42( ) 3 2 3y x x The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to

obtain 4 4– 2( ) 3 2 3y x y x 42 3y x . The result is not the same as

the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.

47. 2 2 12x y Replace x with –x to obtain

2 2 2 2( ) 12 12x y x y . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain

2 2 2 2( ) 12 12x y x y The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin.

48. 2 2 6y x Replace x with –x to obtain

22 2 26 6y x y x


2 2 2 2( ) 6 6y x y x The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and y-axis, it is also symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the x-axis, the y-axis, and the origin.


3 3

3

4( ) ( ) 4( )

4 .

y x x y x xy x x

The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to

obtain 3 34 4 .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain

3 3

3 34( ) ( ) 4( )

4 4 .

y x x y x xy x x y x x

The result is the same as the original equation, so the graph is symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the origin only.




3 3( ) ( ) .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to

obtain 3 3 .y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to

obtain 3 3( ) ( )y x x y x x 3 .y x x The result is the same as the

original equation, so the graph is symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the origin only.

51. 2 8y x x Replace x with –x to obtain

2 2( ) ( ) 8 8.y x x y x x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y

to obtain 2( ) ( ) 8y x x 2 28 8.y x x y x x

The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.

52. y = x + 15 Replace x with –x to obtain

( ) 15 15.y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Because y is a function of x, the graph cannot be symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain ( ) 15 15.y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.

53. 3 2f x x x

3

3 3

2

2 2

f x x xx x x x f x


54. 5 32f x x x

5 3

5 3 5 3

2

2 2

f x x xx x x x f x


55. 4 20.5 2 6f x x x

4 2

4 2

0.5 2 6

0.5 2 6

f x x xx x f x


56. 20.75 4f x x x

2

2

0.75 4

0.75 4

f x x xx x f x


57. 3 9f x x x

3

3 3

9

9 9

f x x xx x x x f x


58. 4 5 8f x x x

4

4

5 8

5 8

f x x xx x f x


59. 2 1f x x


graph of 2y x 1 unit downward.

60. 2 2f x x


graph of 2y x 2 units downward.



61. 2 2f x x



62. 2 3f x x



63. 24g x x



64. 22g x x



65. 22g x x



66. 23g x x



67. 1g x x

The graph is obtained by translating the graph of y x 1 unit downward.

68. 3 2g x x

This graph may be obtained by translating the graph of y x 3 units to the left and 2 units

upward.



69. 3( 1)h x x


graph of 3y x 1 unit to the left. It is then reflected across the x-axis.

70. 3( 1)h x x

This graph can be obtained by translating the

graph of 3y x 1 unit to the right. It is then reflected across the x-axis. (We may also reflect the graph about the x-axis first and then translate it 1 unit to the right.)

71. 22 1h x x


graph of 2y x 1 unit down. It is then stretched vertically by a factor of 2.

72. 23 2h x x


graph of 2y x vertically by a factor of 3, then shifting the resulting graph down 2 units.

73. 22( 2) 4f x x


graph of 2y x 2 units to the right and 4 units down. It is then stretched vertically by a factor of 2.

74. 23( 2) 1f x x


graph of 2y x 2 units to the right and 1 unit up. It is then stretched vertically by a factor of 3 and reflected over the x-axis.

75. 2f x x


graph of y x two units to the left.



76. 3f x x


graph of y x three units to the right.

77. f x x

This graph may be obtained by reflecting the

graph of y x across the x-axis.

78. 2f x x


graph of y x two units down.

79. 2 1f x x


graph of y x vertically by a factor of two and then translating the resulting graph one unit up.

80. 3 2f x x


graph of y x vertically by a factor of three and then translating the resulting graph two units down.

81. 312

4g x x



,

then shifting the resulting graph down four units.

82. 312

2g x x



,

then shifting the resulting graph up two units.



83. 33g x x


graph of 3y x three units left.

84. 32f x x


graph of 3y x two units right.

85. 223

2f x x


graph of 2y x two units to the right, then stretching the resulting graph vertically by a

factor of 23

.

86. Because ( ) ( ),g x x x f x the graphs

are the same.

87. (a) y = g(–x) The graph of g(x) is reflected across the y-axis.

(b) y = g(x – 2) The graph of g(x) is translated to the right 2 units.

(c) y = –g(x) The graph of g(x) is reflected across the x-axis.

(d) y = –g(x) + 2 The graph of g(x) is reflected across the x-axis and translated 2 units up.



88. (a) y f x

The graph of f(x) is reflected across the x-axis.

(b) 2y f x

The graph of f(x) is stretched vertically by a factor of 2.

(c) y f x

The graph of f(x) is reflected across the y-axis.

(d) 12

y f x

The graph of f(x) is compressed vertically

by a factor of 12

.

89. It is the graph of f x x translated 1 unit to

the left, reflected across the x-axis, and translated 3 units up. The equation is

1 3.y x


to the left, reflected across the x-axis, and translated two units up. The equation is

4 2.y x

91. It is the graph of f x x translated one

unit right and then three units down. The

equation is 1 3.y x

92. It is the graph of f x x translated 2 units

to the right, shrunken vertically by a factor of 12

, and translated one unit down. The

equation is 12

2 1.y x


to the left, stretched vertically by a factor of 2, and translated four units down. The equation

is 2 4 4.y x

94. It is the graph of f x x reflected across

the x-axis and then shifted two units down. The equation is 2y x .

95. Because f(3) = 6, the point (3, 6) is on the graph. Because the graph is symmetric with respect to the origin, the point (–3, –6) is on the graph. Therefore, f(–3) = –6.

96. Because f(3) = 6, (3, 6) is a point on the graph. The graph is symmetric with respect to the y-axis, so (–3, 6) is on the graph. Therefore, f(–3) = 6.

97. Because f(3) = 6, the point (3, 6) is on the graph. The graph is symmetric with respect to the line x = 6 and the point (3, 6) is 3 units to the left of the line x = 6, so the image point of (3, 6), 3 units to the right of the line x = 6 is (9, 6). Therefore, f(9) = 6.

98. Because f(3) = 6 and f(–x) = f(x), f(–3) = f(3). Therefore, f(–3) = 6.

99. Because (3, 6) is on the graph, (–3, –6) must also be on the graph. Therefore, f(–3) = –6.

100. If f is an odd function, f(–x) = –f(x). Because f(3) = 6 and f(–x) = –f(x), f(–3) = –f(3). Therefore, f(–3) = –6.

Chapter 2 Quiz (Sections 2.5−2.7) 237


101. f(x) = 2x + 5 Translate the graph of ( )f x up 2 units to obtain the graph of

( ) (2 5) 2 2 7.t x x x Now translate the graph of t(x) = 2x + 7 left 3 units to obtain the graph of

( ) 2( 3) 7 2 6 7 2 13.g x x x x (Note that if the original graph is first translated to the left 3 units and then up 2 units, the final result will be the same.)

102. f(x) = 3 – x Translate the graph of ( )f x down 2 units to

obtain the graph of ( ) (3 ) 2 1.t x x x

Now translate the graph of 1t x x right

3 units to obtain the graph of ( ) ( 3) 1 3 1 4.g x x x x

(Note that if the original graph is first translated to the right 3 units and then down 2 units, the final result will be the same.)

103. (a) Because f(–x) = f(x), the graph is symmetric with respect to the y-axis.

(b) Because f(–x) = –f(x), the graph is symmetric with respect to the origin.

104. (a) f(x) is odd. An odd function has a graph symmetric with respect to the origin. Reflect the left half of the graph in the origin.

(b) f(x) is even. An even function has a graph symmetric with respect to the y-axis. Reflect the left half of the graph in the y-axis.


1. (a) First, find the slope: 9 5

21 ( 3)

m

Choose either point, say, (−3, 5), to find the equation of the line:

5 2( ( 3)) 2( 3) 5y x y x 2 11y x .

(b) To find the x-intercept, let y = 0 and solve

for x: 112

0 2 11x x . The

x-intercept is 112

, 0 .

2. Write 3x − 2y = 6 in slope-intercept form to

find its slope: 32

3 2 6 3.x y y x

Then, the slope of the line perpendicular to

this graph is 23

. 23

4 ( ( 6))y x

2 23 3

( 6)) 4y x y x

3. (a) 8x (b) 5y

4. (a) Cubing function; domain: ( , ) ;


( , ) .

(b) Absolute value function; domain: ( , ) ; range: [0, ) ; decreasing over

( , 0) ; increasing over (0, )



(c) Cube root function: domain: ( , ) ;


( , ) .

5.

0.40 0.75

5.5 0.40 5.5 0.75

0.40 5 0.75 2.75

f x xf

A 5.5-minute call costs $2.75.

6. if 0( )2 3 if 0

x xf xx x

For values of x < 0, the graph is the line y = 2x + 3. Do not include the right endpoint

(0, 3). Graph the line y x for values of x ≥ 0, including the left endpoint (0, 0).

7. 3( ) 1f x x

Reflect the graph of 3( )f x x across the x-axis, and then translate the resulting graph one unit up.

8. ( ) 2 1 3f x x

Shift the graph of ( )f x x one unit right,

stretch the resulting graph vertically by a factor of 2, then shift this graph three units up.

9. This is the graph of ( )g x x , translated four units to the left, reflected across the x-axis, and then translated two units down.

The equation is 4 2y x .

10. (a) 2 7f x x


2

2

( ) 7

7

f x xf x x f x

The result is the same as the original function, so the function is even.

(b) 3 1f x x x


3

3

1

1

f x x xx x f x

The result is not the same as the original equation, so the function is not even. Because ,f x f x the function is

not odd. Therefore, the function is neither even nor odd.

(c) 101 99f x x x


101 99

101 99

101 99

f x x xx x

x xf x

Because f(−x) = −f(x), the function is odd.


In exercises 1–10, 1f x x and 2.g x x

1. 2

2 2 2

2 1 2 7

f g f g

2. 2

2 2 2

2 1 2 1

f g f g

3. 2

2 2 2

2 1 2 12

fg f g

4. 2

2 2 1 32

2 42

ffg g

5. 2 22 2 2 2 1 5f g f g f



6. 22 2 2 1 2 1 9g f g f g

7. f is defined for all real numbers, so its domain is , .

8. g is defined for all real numbers, so its domain is , .

9. f + g is defined for all real numbers, so its domain is , .

10. fg

is defined for all real numbers except those

values that make 0,g x so its domain is

, 0 0, .

In Exercises 11–18, 2( ) 3f x x and

( ) 2 6g x x .

11.

2

( )(3) (3) (3)

(3) 3 2(3) 6

12 0 12

f g f g

12. 2

( )( 5) ( 5) ( 5)

[( 5) 3] [ 2( 5) 6]28 16 44

f g f g

13. 2

( )( 1) ( 1) ( 1)

[( 1) 3] [ 2( 1) 6]4 8 4

f g f g

14. 2

( )(4) (4) (4)

[(4) 3] [ 2(4) 6]19 ( 2) 21

f g f g

15. 2

( )(4) (4) (4)

[4 3] [ 2(4) 6]19 ( 2) 38

fg f g

16. 2

( )( 3) ( 3) ( 3)

[( 3) 3] [ 2( 3) 6]12 12 144

fg f g

17. 2( 1) ( 1) 3 4 1

( 1)( 1) 2( 1) 6 8 2

f fg g

18. 2(5) (5) 3 28

(5) 7(5) 2(5) 6 4

f fg g

19. ( ) 3 4, ( ) 2 5f x x g x x

( )( ) ( ) ( )(3 4) (2 5) 5 1

f g x f x g xx x x

( )( ) ( ) ( )(3 4) (2 5) 9

f g x f x g xx x x

2

2

( )( ) ( ) ( ) (3 4)(2 5)

6 15 8 20

6 7 20

fg x f x g x x xx x xx x

( ) 3 4( )

( ) 2 5

f f x xxg g x x


and fg are all , . The domain of fg is

the set of all real numbers for which 0.g x This is the set of all real numbers

except 52


as 5 52 2

, , .

20. f(x) = 6 – 3x, g(x) = –4x + 1 ( )( ) ( ) ( )

(6 3 ) ( 4 1)7 7

f g x f x g xx x

x

( )( ) ( ) ( )(6 3 ) ( 4 1) 5

f g x f x g xx x x

2

2

( )( ) ( ) ( ) (6 3 )( 4 1)

24 6 12 3

12 27 6

fg x f x g x x xx x x

x x

( ) 6 3( )

( ) 4 1

f f x xxg g x x



0.g x This is the set of all real numbers

except 14


as 1 14 4

– , , .

21. 2 2( ) 2 3 , ( ) 3f x x x g x x x

2 2

2

( )( ) ( ) ( )

(2 3 ) ( 3)

3 4 3


x x

2 2

2 2

2

( )( ) ( ) ( )

(2 3 ) ( 3)

2 3 3

2 3


x x x xx x




(continued)

2 2

4 3 2 3 2

4 3 2

( )( ) ( ) ( )

(2 3 )( 3)

2 2 6 3 3 9

2 5 9 9

fg x f x g xx x x x

x x x x x xx x x x

2

2

( ) 2 3( )

( ) 3

f f x x xxg g x x x



0.g x If 2 3 0x x , then by the

quadratic formula 1 112ix . The equation

has no real solutions. There are no real numbers which make the denominator zero.

Thus, the domain of fg is also , .

22. 2 2( ) 4 2 , ( ) 3 2f x x x g x x x

2 2

2

( )( ) ( ) ( )

(4 2 ) ( 3 2)

5 2


x x

2 2

2 2

2

( )( ) ( ) ( )

(4 2 ) ( 3 2)

4 2 3 2

3 5 2


x x x xx x

2 2

4 3 2 3 2

4 3 2

( )( ) ( ) ( )

(4 2 )( 3 2)

4 12 8 2 6 4

4 10 2 4

fg x f x g xx x x x

x x x x x xx x x x

2

2

( ) 4 2( )

( ) 3 2

f f x x xxg g x x x


and fg are all , . The domain of fg is the

set of all real numbers x such that 2 3 2 0x x . Because 2 3 2 ( 1)( 2)x x x x , the numbers

which give this denominator a value of 0 are

x = 1 and x = 2. Therefore, the domain of fg is

the set of all real numbers except 1 and 2, which is written in interval notation as (– , 1) (1, 2) (2, ) .

23. 1

( ) 4 1, ( )f x x g xx

1( )( ) ( ) ( ) 4 1f g x f x g x x

x

1( )( ) ( ) ( ) 4 1f g x f x g x x

x

( )( ) ( ) ( )

1 4 14 1

fg x f x g xxx

x x

1

( ) 4 1( ) 4 1

( ) x

f f x xx x xg g x

Because 14

4 1 0 4 1 ,x x x the

domain of f is 14





, . Because 1 0x

for any value of x, the domain of fg is also

14

, .

24. 1

( ) 5 4, ( )f x x g xx

( )( ) ( ) ( )1 1

5 4 5 4

f g x f x g x

x xx x

( )( ) ( ) ( )1 1

5 4 5 4

f g x f x g x

x xx x

( )( ) ( ) ( )

1 5 45 4

fg x f x g xxx

x x

1

( ) 5 4( ) 5 4

( ) x

f f x xx x xg g x

Because 45

5 4 0 5 4 ,x x x the

domain of f is 45





, . 1 0x for any

value of x, so the domain of fg is also

45

, .



25. 2008 280 and 2008 470, thus

2008 2008 2008280 470 750 (thousand).

M FT M F

26. 2012 390 and 2012 630, thus

2012 2012 2012390 630 1020 (thousand).

M FT M F

27. Looking at the graphs of the functions, the slopes of the line segments for the period 2008−2012 are much steeper than the slopes of the corresponding line segments for the period 2004−2008. Thus, the number of associate’s degrees increased more rapidly during the period 2008−2012.

28. If 2004 2012, ( ) ,k T k r and F(k) = s, then M(k) = r − s.

29. 2000 2000 200019 13 6

T S T S

It represents the dollars in billions spent for general science in 2000.

30. 2010 2010 201029 11 18

T G T G

It represents the dollars in billions spent on space and other technologies in 2010.

31. Spending for space and other technologies spending decreased in the years 1995−2000 and 2010–2015.

32. Total spending increased the most during the years 2005−2010.

33. (a)

2 2 2

4 2 2

f g f g

(b) ( )(1) (1) (1) 1 ( 3) 4f g f g

(c) ( )(0) (0) (0) 0( 4) 0fg f g

(d) (1) 1 1

(1)(1) 3 3

f fg g

34. (a) ( )(0) (0) (0) 0 2 2f g f g

(b) ( )( 1) ( 1) ( 1)2 1 3

f g f g

(c) ( )(1) (1) (1) 2 1 2fg f g

(d) (2) 4

(2) 2(2) 2

f fg g

35. (a) ( )( 1) ( 1) ( 1) 0 3 3f g f g

(b) ( )( 2) ( 2) ( 2)1 4 5

f g f g

(c) ( )(0) (0) (0) 1 2 2fg f g

(d) (2) 3

(2) undefined(2) 0

f fg g

36. (a) ( )(1) (1) (1) 3 1 2f g f g

(b) ( )(0) (0) (0) 2 0 2f g f g

(c) ( )( 1) ( 1) ( 1) 3( 1) 3fg f g

(d) (1) 3

(1) 3(1) 1

f fg g

37. (a) 2 2 2 7 2 5f g f g

(b) ( )(4) (4) (4) 10 5 5f g f g

(c) ( )( 2) ( 2) ( 2) 0 6 0fg f g

(d) (0) 5

(0) undefined(0) 0

f fg g

38. (a) 2 2 2 5 4 9f g f g

(b) ( )(4) (4) (4) 0 0 0f g f g

(c) ( )( 2) ( 2) ( 2) 4 2 8fg f g

(d) (0) 8

(0) 8(0) 1

f fg g




−2 0 6 0 6 6 0 6 6 0 6 0 06

0

0 5 0 5 0 5 5 0 5 5 0 0 50

undefined

2 7 −2 7 2 5 7 2 9 7 2 14 72

3.5

4 10 5 10 5 15 10 5 5 10 5 50 105

2


−2 −4 2 4 2 2 4 2 6 4 2 8 42

2

0 8 −1 8 1 7 8 1 9 8 1 8 81

8

2 5 4 5 4 9 5 4 1 5 4 20 54

1.25

4 0 0 0 0 0 0 0 0 0 0 0 00

undefined

41. Answers may vary. Sample answer: Both the

slope formula and the difference quotient represent the ratio of the vertical change to the horizontal change. The slope formula is stated for a line while the difference quotient is stated for a function f.

42. Answers may vary. Sample answer: As h approaches 0, the slope of the secant line PQ approaches the slope of the line tangent of the curve at P.

43. 2f x x

(a) ( ) 2 ( ) 2f x h x h x h

(b) ( ) ( ) (2 ) (2 )2 2


(c) ( ) ( )

1f x h f x h

h h

44. 1f x x

(a) ( ) 1 ( ) 1f x h x h x h

(b) ( ) ( ) (1 ) (1 )1 1


(c) ( ) ( )

1f x h f x h

h h

45. 6 2f x x

(a) ( ) 6( ) 2 6 6 2f x h x h x h

(b) ( ) ( )(6 6 2) (6 2)6 6 2 6 2 6

f x h f xx h x

x h x h

(c) ( ) ( ) 6

6f x h f x h

h h

46. 4 11f x x

(a) ( ) 4( ) 11 4 4 11f x h x h x h

(b) ( ) ( )(4 4 11) (4 11)4 4 11 4 11 4

f x h f xx h x

x h x h

(c) ( ) ( ) 4

4f x h f x h

h h

47. 2 5f x x

(a) ( ) 2( ) 52 2 5

f x h x hx h

(b) ( ) ( )( 2 2 5) ( 2 5)

2 2 5 2 5 2

f x h f xx h x

x h x h

(c) ( ) ( ) 2

2f x h f x h

h h



48. ( ) 4 2f x x

(a) ( ) 4( ) 24 4 2

f x h x hx h

(b)

( ) ( )4 4 2 4 24 4 2 4 24

f x h f xx h xx h xh

(c) ( ) ( ) 4

4f x h f x h

h h

49. 1

( )f xx

(a) 1

( )f x hx h

(b)

( ) ( )

1 1

f x h f xx x h

x h x x x hh

x x h

(c)

( ) ( )

1

hx x hf x h f x h

h h hx x h

x x h

50. 2

1( )f x

x

(a) 2

1( )f x h

x h

(b)

22

2 2 22

2 2 2 2

2 22 2

( ) ( )

1 1

2 2

f x h f xx x h

xx h x x hx x xh h xh h

x x h x x h

(c)

2

22

22

22

22

22

( ) ( ) 2

2

2

xh hx x hf x h f x xh h

h h hx x hh x h

hx x hx h

x x h

51. 2( )f x x

(a) 2 2 2( ) ( ) 2f x h x h x xh h

(b) 2 2 2

2( ) ( ) 2

2

f x h f x x xh h xxh h

(c) 2( ) ( ) 2

(2 )

2

f x h f x xh hh h

h x hh

x h

52. 2( )f x x

(a)

2

2 2

2 2

( ) ( )

2

2

f x h x hx xh h

x xh h

(b) 2 2 2

2 2 2

2

( ) ( ) 2

2

2

f x h f x x xh h x

x xh h xxh h

(c) 2( ) ( ) 2

(2 )

2

f x h f x xh hh h

h x hh

x h

53. 2( ) 1f x x

(a) 2

2 2

2 2

( ) 1 ( )

1 ( 2 )

1 2

f x h x hx xh h

x xh h

(b) 2 2 2

2 2 2

2

( ) ( )

(1 2 ) (1 )

1 2 1

2

f x h f xx xh h x

x xh h xxh h

(c) 2( ) ( ) 2

( 2 )

2

f x h f x xh hh h

h x hh

x h

54. 2( ) 1 2f x x

(a) 2

2 2

2 2

( ) 1 2( )

1 2( 2 )

1 2 4 2

f x h x hx xh h

x xh h



(b)

2 2 2

2 2 2

2

( ) ( )

1 2 4 2 1 2

1 2 4 2 1 2

4 2

f x h f xx xh h x

x xh h xxh h

(c) 2( ) ( ) 4 2

(4 2 )

4 2

f x h f x xh hh h

h x hh

x h

55. 2( ) 3 1f x x x

(a) 2

2 2

( ) 3 1

2 3 3 1


(b)

2 2

2

2 2 2

2

( ) ( )

2 3 3 1

3 1

2 3 3 1 3 1

2 3

f x h f xx xh h x h

x x


(c) 2( ) ( ) 2 3

(2 3)

2 3

f x h f x xh h hh h

h x hh

x h

56. 2( ) 4 2f x x x

(a) 2

2 2

( ) 4 2

2 4 4 2


(b)

2 2

2

2 2 2

2

( ) ( )

2 4 4 2

4 2

2 4 4 2 4 2

2 4

f x h f xx xh h x h

x x


(c) 2( ) ( ) 2 4

(2 4)

2 4

f x h f x xh h hh h

h x hh

x h

57. 3 4 4 3 1g x x g

4 4 1

2 1 3 2 3 5

f g f g f

58. 3 2 2 3 1g x x g

2 2 1

2 1 3 2 3 1

f g f g f

59. 3 2 2 3 5g x x g

2 2 5

2 5 3 10 3 7

f g f g f

60. 2 3f x x 3 2 3 3 6 3 3f

3 3 3 3 3 0g f g f g

61. 2 3 0 2 0 3 0 3 3f x x f

0 0 3

3 3 3 3 6

g f g f g

62. 2 3 2 2 2 3 7f x x f

2 2 7

7 3 7 3 10

g f g f g

63. 2 3 2 2 2 3 4 3 1f x x f

2 2 1 2 1 3 1f f f f f

64. 3 2 2 3 5g x x g

2 2 5 5 3 2g g g g g

65. ( )(2) [ (2)] (3) 1f g f g f

66. ( )(7) [ (7)] (6) 9f g f g f

67. ( )(3) [ (3)] (1) 9g f g f g

68. ( )(6) [ (6)] (9) 12g f g f g

69. ( )(4) [ (4)] (3) 1f f f f f

70. ( )(1) [ (1)] (9) 12g g g g g

71. ( )(1) [ (1)] (9)f g f g f However, f(9) cannot be determined from the table given.

72. ( ( ))(7) ( ( (7)))( (6)) (9) 12

g f g g f gg f g

73. (a) ( )( ) ( ( )) (5 7)6(5 7) 930 42 9 30 33

f g x f g x f xx

x x


( , ) .



(b) ( )( ) ( ( )) ( 6 9)5( 6 9) 7

30 45 7 30 52

g f x g f x g xx

x x


74. (a) ( )( ) ( ( )) (3 1)8(3 1) 1224 8 12 24 4

f g x f g x f xxx x


( , ) .

(b) ( )( ) ( ( )) (8 12)3(8 12) 124 36 1 24 35

g f x g f x g xxx x


75. (a) ( )( ) ( ( )) ( 3) 3f g x f g x f x x

The domain and range of g are ( , ) , however, the domain and range of f are [0, ) . So, 3 0 3x x .


[ 3, ) .

(b) ( )( ) ( ( )) 3g f x g f x g x x

The domain and range of g are ( , ) , however, the domain and range of f are [0, ) .Therefore, the domain of g f is

[0, ) .

76. (a) ( )( ) ( ( )) ( 1) 1f g x f g x f x x

The domain and range of g are ( , ) , however, the domain and range of f are [0, ) . So, 1 0 1x x . Therefore,

the domain of f g is [1, ) .

(b) ( )( ) ( ( )) 1g f x g f x g x x

The domain and range of g are ( , ), however, the domain and range of f are [0, ). Therefore, the domain of g f is

[0, ).

77. (a) 2

2 3

( )( ) ( ( )) ( 3 1)

( 3 1)

f g x f g x f x xx x

The domain and range of f and g are ( , ), so the domain of f g is

( , ).

(b)

3

23 3

6 3

( )( ) ( ( )) ( )

3 1

3 1

g f x g f x g x

x x

x x

The domain and range of f and g are ( , ), so the domain of g f is

( , ).

78. (a) 4 2

4 2

4 2

( )( ) ( ( )) ( 4)

4 2

2

f g x f g x f x xx xx x



g is 4, , so the domain of f g is

( , ).

(b) 4 2

( )( ) ( ( )) ( 2)

( 2) ( 2) 4

g f x g f x g xx x



g is 4, , so the domain of g f is

( , ).

79. (a) ( )( ) ( ( )) (3 ) 3 1f g x f g x f x x


however, the domain of f is [1, ), while

the range of f is [0, ). So, 13

3 1 0x x . Therefore, the

domain of f g is 13

, .

(b) ( )( ) ( ( )) 1

3 1

g f x g f x g x

x




80. (a) ( )( ) ( ( )) (2 ) 2 2f g x f g x f x x



2 2 0 1x x . Therefore, the




(b) ( )( ) ( ( )) 2g f x g f x g x

2 2x




81. (a) 21

( )( ) ( ( )) ( 1) xf g x f g x f x



( , 1) ( 1, ) .

(b) 2 2( )( ) ( ( )) 1x xg f x g f x g




( ,0) (0, ) .

82. (a) 44

( )( ) ( ( )) ( 4) xf g x f g x f x



( , 4) ( 4, ) .

(b) 4 4( )( ) ( ( )) 4x xg f x g f x g




( ,0) (0, ) .

83. (a) 1 1( )( ) ( ( )) 2x xf g x f g x f


of f is [ 2, ) . So, 1 2 0x

12



12

, 0 , .

(b) 12

( )( ) ( ( )) 2x

g f x g f x g x



are ( ,0) (0, ) . So 2 0 2x x .


84. (a) 2 2( )( ) ( ( )) 4x xf g x f g x f


of f is [ 4, ) . So, 2 4 0x

12



12

, 0 , .

(b) 24

( )( ) ( ( )) 4x

g f x g f x g x



are ( , 0) (0, ) . So 4 0 4x x .


85. (a) 1 15 5

( )( ) ( ( )) x xf g x f g x f





(b) 15

( )( ) ( ( ))x

g f x g f x g x

The domain and range of f is [0, ). The

domain of g is ( , 5) ( 5, ).


86. (a) 3 36 6

( )( ) ( ( )) x xf g x f g x f





(b) 36

( )( ) ( ( ))x

g f x g f x g x

The domain and range of f is [0, ) . The

domain of g is ( , 6) ( 6, ) .




87. (a) 1 11 2 1 2

( )( ) ( ( )) xx x xf g x f g x f

The domain and range of g are ( , 0) (0, ) . The domain of f is


( , 0) (0, ) . So, 1 2

0 0xx x or

12

0 x or 12

x (using test intervals).

Thus, 0x and 12

x . Therefore, the

domain of f g is

1 12 2

, 0 0, , .

(b) 1 12 1 ( 2)

( )( ) ( ( ))

2x xg f x g f x g

x


( , 2) (2, ), and the range of f is

( , 0) (0, ). Therefore, the domain of

g f is ( , 2) (2, ).

88. (a) 1 11 4

1 4

( )( ) ( ( )) x xx

x

f g x f g x f



( , 0) (0, ). So, 1 4

0 0xx x

or 14

0 x or 14

1 4 0x x

(using test intervals). Thus, x ≠ 0 and

x ≠ 14

. Therefore, the domain of f g is

1 14 4

, 0 0, , .

(b) 1 14 1 ( 4)

( )( ) ( ( ))

4x xg f x g f x g

x

The domain and range of g are ( ,0) (0, ) . The domain of f is

( , 4) ( 4, ) , and the range of f is

( ,0) (0, ) . Therefore, the domain of

g f is ( , 4) ( 4, ).

89.

(2) (1) 2 and (3) (2) 5

Since (1) 7 and (1) 3, (3) 7.

g f g g f gg f f g

x f x g x ( )g f x

1 3 2 7

2 1 5 2

3 2 7 5

90. ( )f x is odd, so ( 1) (1) ( 2) 2.f f

Because ( )g x is even, (1) ( 1) 2g g and

(2) ( 2) 0.g g ( )( 1) 1,f g so

( 1) 1f g and (2) 1.f ( )f x is odd, so

( 2) (2) 1.f f Thus,

( )( 2) ( 2) (0) 0f g f g f and

( )(1) (1) (2) 1f g f g f and

( )(2) (2) (0) 0.f g f g f

x –2 –1 0 1 2

f x –1 2 0 –2 1

g x 0 2 1 2 0

f g x 0 1 –2 1 0

91. Answers will vary. In general, composition of functions is not commutative. Sample answer:

2 3 3 2 3 26 9 2 6 11

3 2 2 3 2 36 4 3 6 7

f g x f x xx x

g f x g x xx x

Thus, f g x is not equivalent to

g f x .

92.

3

33

7

7 7

7 7

f g x f g x f x

xx x

3

33 33

7

7 7

g f x g f x g x

x x x

93.

14

14

4 2 2

4 2 2

2 2 2 2

f g x f g x x

xx x x

14

1 14 4

4 2 2

4 2 2 4

g f x g f x xx x x

94.

13

13

3

3

f g x f g x x

x x

13

13

3

3

g f x g f x x

x x



95. 31 435 5

3 33 3

5 4

4 4

f g x f g x x

x x x

31 435 5

51 4 4 45 5 5 5 555

5 4

5 4 x

x

g f x g f x x

xx

96. 33

3 33 3

1 1

1 1

f g x f g x x

x x x

33 1 1

1 1

g f x g f x xx x

In Exercises 97−102, we give only one of many possible answers.

97. 2( ) (6 2)h x x

Let g(x) = 6x – 2 and 2( )f x x . 2( )( ) (6 2) (6 2) ( )f g x f x x h x

98. 2 2( ) (11 12 )h x x x 2 2Let ( ) 11 12 and ( ) .g x x x f x x

2

2 2

( )( ) (11 12 )

(11 12 ) ( )

f g x f x xx x h x

99. 2( ) 1h x x

Let 2( ) 1 and ( ) .g x x f x x

2 2( )( ) ( 1) 1 ( ).f g x f x x h x

100. 3( ) (2 3)h x x 3Let ( ) 2 3 and ( ) .g x x f x x

3( )( ) (2 3) (2 3) ( )f g x f x x h x

101. ( ) 6 12h x x

Let ( ) 6 and ( ) 12.g x x f x x

( )( ) (6 ) 6 12 ( )f g x f x x h x

102. 3( ) 2 3 4h x x

Let 3( ) 2 3 and ( ) 4.g x x f x x 3( )( ) (2 3) 2 3 4 ( )f g x f x x h x

103. f(x) = 12x, g(x) = 5280x ( )( ) [ ( )] (5280 )

12(5280 ) 63,360f g x f g x f x

x x

The function f g computes the number of inches in x miles.

104. 3 , 1760f x x g x x

1760

3 1760 5280

f g x f g x f xx x

f g x compute the number of feet in x

miles.

105. 23( )

4x x

(a) 2 2 23 3(2 ) (2 ) (4 ) 3

4 4x x x x

(b) 2(16) (2 8) 3(8)

64 3 square units

A

106. (a) 4 4

4 x xx s s s

(b) 2224 16x xy s

(c) 26 36

2.25 square units16 16

y

107. (a) 2( ) 4 and ( )r t t r r

2 2( )( ) [ ( )]

(4 ) (4 ) 16

r t r tt t t

(b) ( )( )r t defines the area of the leak in terms of the time t, in minutes.

(c) 2 2(3) 16 (3) 144 ft

108. (a) 2 2

( )( ) [ ( )]

(2 ) (2 ) 4

r t r tt t t

(b) It defines the area of the circular layer in terms of the time t, in hours.

(c) 2 2( )(4) 4 (4) 64 mir

109. Let x = the number of people less than 100 people that attend.

(a) x people fewer than 100 attend, so 100 – x people do attend N(x) = 100 – x

(b) The cost per person starts at $20 and increases by $5 for each of the x people that do not attend. The total increase is $5x, and the cost per person increases to $20 + $5x. Thus, G(x) = 20 + 5x.

(c) ( ) ( ) ( ) (100 )(20 5 )C x N x G x x x



(d) If 80 people attend, x = 100 – 80 = 20.

20 100 20 20 5 20

80 20 100

80 120 $9600

C

110. (a) 1 0.02y x

(b) 2 0.015( 500)y x

(c) 1 2y y represents the total annual interest.

(d) 1 2

1 2

( )(250)(250) (250)

0.02(250) 0.015(250 500)5 0.015(750) 15 11.25$16.25

y yy y

111. (a) 1

2g x x

(b) 1f x x

(c) 1 11

2 2f g x f g x f x x

(d) 160 60 1 $31

2f g

112. If the area of a square is 2x square inches, each side must have a length of x inches. If 3 inches is added to one dimension and 1 inch is subtracted from the other, the new dimensions will be x + 3 and x – 1. Thus, the area of the resulting rectangle is (x) = (x + 3)(x – 1).

Chapter 2 Review Exercises

1. P(3, –1), Q(–4, 5) 2 2

2 2

( , ) ( 4 3) [5 ( 1)]

( 7) 6 49 36 85

d P Q

Midpoint:

3 ( 4) 1 5 1 4 1, , , 2

2 2 2 2 2

2. M(–8, 2), N(3, –7) 2 2

2 2

( , ) [3 ( 8)] ( 7 2)

11 ( 9) 121 81 202

d M N

Midpoint: 8 3 2 ( 7) 5 5

, , 2 2 2 2

3. A(–6, 3), B(–6, 8) 2 2

2

( , ) [ 6 ( 6)] (8 3)

0 5 25 5

d A B

Midpoint: 6 ( 6) 3 8 12 11 11

, , 6,2 2 2 2 2

4. Label the points A(5, 7), B(3, 9), and C(6, 8).

2 2

2 2

( , ) 3 5 9 7

2 2 4 4 8

d A B

2 2

2 2

( , ) (6 5) (8 7)

1 1 1 1 2

d A C

2 2

22

( , ) 6 3 8 9

3 1 9 1 10

d B C


ABC is a right triangle with right angle at (5, 7).

5. Let B have coordinates (x, y). Using the midpoint formula, we have

6 108, 2 ,

2 26 10

8 22 2

6 16 10 422 6

x y

x y

x yx y

The coordinates of B are (22, −6).

6. P(–2, –5), Q(1, 7), R(3, 15) 2 2

2 2

( , ) (1 ( 2)) (7 ( 5))

(3) (12) 9 144

153 3 17

d P Q

2 2

2 2

( , ) (3 1) (15 7)

2 8 4 64 68 2 17

d Q R

2 2

2 2

( , ) (3 ( 2)) (15 ( 5))

(5) (20) 25 400 5 17

d P R




(continued)

( , ) ( , ) 3 17 2 17d P Q d Q R

5 17 ( , ),d P R so these three points are collinear.

7. Center (–2, 3), radius 15 2 2 2

2 2 2

2 2

( ) ( )

[ ( 2)] ( 3) 15

( 2) ( 3) 225

x h y k rx y

x y

8. Center ( 5, 7 ), radius 3

2 2 2

22 2

2 2

( ) ( )

5 7 3

5 7 3

x h y k r

x y

x y

9. Center (–8, 1), passing through (0, 16) The radius is the distance from the center to any point on the circle. The distance between (–8, 1) and (0, 16) is

2 2 2 2(0 ( 8)) (16 1) 8 15

64 225 289 17.

r


2 2[ ( 8)] ( 1) 17

( 8) ( 1) 289

x yx y

10. Center (3, –6), tangent to the x-axis The point (3, –6) is 6 units directly below the x-axis. Any segment joining a circle’s center to a point on the circle must be a radius, so in this case the length of the radius is 6 units.

2 2 2

2 2 2

2 2

( ) ( )

( 3) [ ( 6)] 6

( 3) ( 6) 36

x h y k rx y

x y


2 2 2(3 0) (5 0) 9 25 34r



2 2 2( 2 0) (3 0) 4 9 13r



2 2 2( 2 0) (6 3) 4 9 13r

The equation is 2 2( 3) 13x y .


2 2 2(4 5) (9 6) 1 9 10r

The equation is 2 2( 5) ( 6) 10x y .


2 2

2 2

2 2

4 6 12

4 4 6 9 12 4 9

2 3 1

x x y y

x x y y

x y

The circle has center (2, –3) and radius 1.


2 2

2 2

( 6 9) ( 10 25) 30 9 25

( 3) ( 5) 4

x x y yx y

The circle has center (3, 5) and radius 2.

17.

2 2

2 2

2 2

2 249 9 49 94 4 4 4

2 27 3 49 942 2 4 4 4

2 27 3 542 2 4

2 14 2 6 2 0

7 3 1 0

7 3 1

7 3 1

x x y yx x y yx x y y

x x y y

x y

x y

The circle has center 7 32 2

, and radius

54 9 6 3 6544 24 4

.

18.

2 2

2 2

2 2

2 2 25 25121 1214 4 4 4

22 5 146112 2 4

3 33 3 15 0

11 5 0

11 5 0

11 5 0

x x y yx x y y

x x y y

x x y y

x y

The circle has center 5112 2

, and radius

1462

.

19. This is not the graph of a function because a vertical line can intersect it in two points. domain: [−6, 6]; range: [−6, 6]

20. This is not the graph of a function because a vertical line can intersect it in two points. domain: , ; range: 0,



21. This is not the graph of a function because a vertical line can intersect it in two points. domain: , ; range: ( , 1] [1, )

22. This is the graph of a function. No vertical line will intersect the graph in more than one point. domain: , ; range: 0,

23. This is not the graph of a function because a vertical line can intersect it in two points. domain: 0, ; range: ,

24. This is the graph of a function. No vertical line will intersect the graph in more than one point. domain: , ; range: ,

25. 26y x Each value of x corresponds to exactly one value of y, so this equation defines a function.

26. The equation 213

x y does not define y as a

function of x. For some values of x, there will be more than one value of y. For example, ordered pairs (3, 3) and (3, –3) satisfy the relation. Thus, the relation would not be a function.

27. The equation 2y x does not define y as a function of x. For some values of x, there will be more than one value of y. For example, ordered pairs (3, 1) and (3, −1) satisfy the relation.

28. The equation 4yx

defines y as a function

of x because for every x in the domain, which is (– , 0) (0, ) , there will be exactly one value of y.

29. In the function ( ) 4f x x , we may use

any real number for x. The domain is , .

30. 8

( )8

xf xx

x can be any real number except 8 because this will give a denominator of zero. Thus, the domain is (– , 8) (8, ).

31. 6 3f x x

In the function 6 3y x , we must have

6 3 0x . 6 3 0 6 3 2 2x x x x Thus, the domain is , 2 .

32. (a) As x is getting larger on the interval 2, , the value of y is increasing.

(b) As x is getting larger on the interval , 2 , the value of y is decreasing.

(c) f(x) is constant on (−2, 2).

In exercises 33–36, 22 3 6.f x x x

33. 23 2 3 3 3 62 9 3 3 618 9 6 15

f

34.

20.5 2 0.5 3 0.5 6

2 0.25 3 0.5 60.5 1.5 6 8

f

35. 20 2 0 3 0 6 6f

36. 22 3 6f k k k

37. 25

2 5 5 5 2 5 1x y y x y x


and

y-intercept (0, –)1. It may also be graphed using intercepts. To do this, locate the x-intercept: 0y

52

2 5 0 5 2 5x x x

38. 37

3 7 14 7 3 14 2x y y x y x


and

y-intercept (0, 2). It may also be graphed using intercepts. To do this, locate the x-intercept by setting y = 0:

143

3 7 0 14 3 14x x x



39. 25

2 5 20 5 2 20 4x y y x y x


and

y-intercept (0, 4). It may also be graphed using intercepts. To do this, locate the x-intercept: x-intercept: 0y

2 5 0 20 2 20 10x x x

40. 13

3y x y x


and

y-intercept (0, 0), which means that it passes through the origin. Use another point such as (6, 2) to complete the graph.

41. f(x) = x The graph is the line with slope 1 and y-intercept (0, 0), which means that it passes through the origin. Use another point such as (1, 1) to complete the graph.

42.

14

4 84 8

2

x yy xy x


and

y-intercept (0, –2). It may also be graphed using intercepts. To do this, locate the x-intercept:

0 4 0 8 8y x x

43. x = –5 The graph is the vertical line through (–5, 0).

44. f(x) = 3 The graph is the horizontal line through (0, 3).

45. 2 0 2y y The graph is the horizontal line through (0, −2).

46. The equation of the line that lies along the x-axis is y = 0.

47. Line through (0, 5), 23

m

Note that 2 23 3

.m

Begin by locating the point (0, 5). Because the

slope is 23 , a change of 3 units horizontally

(3 units to the right) produces a change of –2 units vertically (2 units down). This gives a second point, (3, 3), which can be used to complete the graph.




(continued)

48. Line through (2, –4), 34

m

First locate the point (2, –4).

Because the slope is 34

, a change of 4 units

horizontally (4 units to the right) produces a change of 3 units vertically (3 units up). This gives a second point, (6, –1), which can be used to complete the graph.

49. through (2, –2) and (3, –4) 2 1

2 1

4 2 22

3 2 1

y ymx x

50. through (8, 7) and 12

, 2

2 11 15

2 1 2 2

2 7 9

8

2 18 69

15 15 5

y ymx x

51. through (0, –7) and (3, –7) 7 7 0

03 0 3

m

52. through (5, 6) and (5, –2)

2 1

2 1

2 6 8

5 5 0

y ymx x

The slope is undefined.

53. 11 2 3x y Solve for y to put the equation in slope-intercept form.

3112 2

2 11 3y x y x


.

54. 9x – 4y = 2. Solve for y to put the equation in slope-intercept form.

9 14 2

4 9 2y x y x


.

55. 2 0 2x x The graph is a vertical line, through (2, 0). The slope is undefined.

56. x – 5y = 0. Solve for y to put the equation in slope-intercept form.

15

5y x y x


.

57. Initially, the car is at home. After traveling for 30 mph for 1 hr, the car is 30 mi away from home. During the second hour the car travels 20 mph until it is 50 mi away. During the third hour the car travels toward home at 30 mph until it is 20 mi away. During the fourth hour the car travels away from home at 40 mph until it is 60 mi away from home. During the last hour, the car travels 60 mi at 60 mph until it arrived home.

58. (a) This is the graph of a function because no vertical line intersects the graph in more than one point.

(b) The lowest point on the graph occurs in December, so the most jobs lost occurred in December. The highest point on the graph occurs in January, so the most jobs gained occurred in January.

(c) The number of jobs lost in December is approximately 6000. The number of jobs gained in January is approximately 2000.

(d) It shows a slight downward trend.

59. (a) We need to first find the slope of a line that passes between points (0, 30.7) and (12, 82.9)

2 1

2 1

82.9 30.7 52.24.35

12 0 12

y ymx x

Now use the point-intercept form with b = 30.7 and m = 4.35. y = 4.35x + 30.7 The slope, 4.35, indicates that the number of e-filing taxpayers increased by 4.35% each year from 2001 to 2013.

(b) For 2009, we evaluate the function for x = 8. y = 4.35(8) + 30.7 = 65.5 65.5% of the tax returns are predicted to have been filed electronically.



60. We need to find the slope of a line that passes between points (1980, 21000) and (2013, 63800)

2 1

2 1

63,800 21,000

2013 198042,800

$1297 per year33

y ymx x

The average rate of change was about $1297 per year.

61. (a) through (3, –5) with slope –2 Use the point-slope form.

1 1( )( 5) 2( 3)

5 2( 3)5 2 6

2 1

y y m x xy x

y xy x

y x

(b) Standard form: 2 1 2 1y x x y

62. (a) through (–2, 4) and (1, 3) First find the slope.

3 4 1

1 ( 2) 3m

Now use the point-slope form with 1

1 1 3( , ) (1, 3) and .x y m

1 113

1013 3

( )

3 ( 1)3( 3) 1( 1)

3 9 13 10

y y m x xy x

y xy x

y x y x


3 33 10

3 10y x y xx y

63. (a) through (2, –1) parallel to 3x – y = 1 Find the slope of 3x – y = 1. 3 1 3 1 3 1x y y x y x The slope of this line is 3. Because parallel lines have the same slope, 3 is also the slope of the line whose equation is to be found. Now use the point-slope form with 1 1( , ) (2, 1) and 3.x y m

1 1( )( 1) 3( 2)

1 3 6 3 7

y y m x xy x

y x y x

(b) Standard form: 3 7 3 7 3 7y x x y x y

64. (a) x-intercept (–3, 0), y-intercept (0, 5) Two points of the line are (–3, 0) and (0, 5). First, find the slope.

5 0 5

0 3 3m

The slope is 53

and the y-intercept is

(0, 5). Write the equation in slope-

intercept form: 53

5y x


5 3 5 155 3 15 5 3 15

y x y xx y x y

65. (a) through (2, –10), perpendicular to a line with an undefined slope A line with an undefined slope is a vertical line. Any line perpendicular to a vertical line is a horizontal line, with an equation of the form y = b. The line passes through (2, –10), so the equation of the line is y = –10.


66. (a) through (0, 5), perpendicular to 8x + 5y = 3 Find the slope of 8x + 5y = 3.

8 35 5

8 5 3 5 8 3x y y xy x

The slope of this line is 85

. The slope

of any line perpendicular to this line is 58

, because 8 55 8

1.

The equation in slope-intercept form with

slope 58

and y-intercept (0, 5) is

58

5.y x


5 8 5 405 8 40 5 8 40

y x y xx y x y

67. (a) through (–7, 4), perpendicular to y = 8 The line y = 8 is a horizontal line, so any line perpendicular to it will be a vertical line. Because x has the same value at all points on the line, the equation is x = –7. It is not possible to write this in slope-intercept form.

(b) Standard form: x = –7

68. (a) through (3, –5), parallel to y = 4 This will be a horizontal line through (3, –5). Because y has the same value at all points on the line, the equation is y = –5.




69. ( ) 3f x x

The graph is the same as that of ,y x

except that it is translated 3 units downward.

70. ( )f x x

The graph of ( )f x x is the reflection of

the graph of y x about the x-axis.

71. 2( ) 1 3f x x

The graph of 2( ) 1 3f x x is a

translation of the graph of 2y x to the left 1 unit, reflected over the x-axis and translated up 3 units.

72. ( ) 2f x x

The graph of ( ) 2f x x is the reflection

of the graph of y x about the x-axis, translated down 2 units.

73. ( ) 3f x x

To get y = 0, we need 0 3 1x 3 4.x To get y = 1, we need1 3 2x 4 5.x Follow this pattern to graph the step function.

74. 3( ) 2 1 2f x x

The graph of 3( ) 2 1 2f x x is a

translation of the graph of 3y x to the left 1 unit, stretched vertically by a factor of 2, and translated down 2 units.

75. 4 2 if 1( )

3 5 if 1x xf x x x

Draw the graph of y = –4x + 2 to the left of x = 1, including the endpoint at x = 1. Draw the graph of y = 3x – 5 to the right of x = 1, but do not include the endpoint at x = 1. Observe that the endpoints of the two pieces coincide.



76. 2 3 if 2( )

4 if 2x xf x

x x

Graph the curve 2 3y x to the left of x = 2, and graph the line y = –x + 4 to the right of x = 2. The graph has an open circle at (2, 7) and a closed circle at (2, 2).

77. if 3

( )6 if 3x xf x

x x

Draw the graph of y x to the left of x = 3,

but do not include the endpoint. Draw the graph of y = 6 – x to the right of x = 3, including the endpoint. Observe that the endpoints of the two pieces coincide.

78. Because x represents an integer, .x x

Therefore, 2 .x x x x x

79. True. The graph of an even function is symmetric with respect to the y-axis.

80. True. The graph of a nonzero function cannot be symmetric with respect to the x-axis. Such a graph would fail the vertical line test

81. False. For example, 2( )f x x is even and (2, 4) is on the graph but (2, –4) is not.

82. True. The graph of an odd function is symmetric with respect to the origin.

83. True. The constant function 0f x is both

even and odd. Because 0 ,f x f x

the function is even. Also 0 0 ,f x f x so the function is

odd.

84. False. For example, 3( )f x x is odd, and (2, 8) is on the graph but (–2, 8) is not.

85. 2 10x y

Replace x with –x to obtain 2( ) 10.x y The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to

obtain 2 2( ) 10 10.x y x y The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. Replace x with –x and y with –y to

obtain 2 2( ) ( ) 10 ( ) 10.x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. The graph is symmetric with respect to the x-axis only.

86. 2 25 5 30y x Replace x with –x to obtain

2 2 2 25 5( ) 30 5 5 30.y x y x The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain

2 2 2 25( ) 5 30 5 5 30.y x y x The result is the same as the original equation, so the graph is symmetric with respect to the x-axis. The graph is symmetric with respect to the y-axis and x-axis, so it must also be symmetric with respect to the origin. Note that

this equation is the same as 2 2 6y x , which is a circle centered at the origin.

87. 2 3x y Replace x with –x to obtain

2 3 2 3( ) .x y x y The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace

y with –y to obtain 2 3 2 3( ) .x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y

with –y to obtain 2 3 2 3( ) ( ) .x y x y The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-axis only.



88. 3 4y x

Replace x with –x to obtain 3 4y x . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain

3 3( ) 4 4y x y x 3 4y x The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain

3 3 3( ) ( ) 4 4 4.y x y x y x The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.

89. 6 4x y


6 4x y . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain 6 ( ) 4 6 4x y x y . The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain 6( ) ( ) 4x y 6 4x y . This equation is not equivalent to the original one, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries.

90. y x

Replace x with –x to obtain ( ) .y x y x The result is not the

same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain

.y x y x The result is the same as

the original equation, so the graph is symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain

( ) .y x y x The result is not the

same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the x-axis only.

91. y = 1 This is the graph of a horizontal line through (0, 1). It is symmetric with respect to the y-axis, but not symmetric with respect to the x-axis and the origin.

92. x y

Replace x with –x to obtain .x y x y


.x y x y The result is the same as

the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and with respect to the y-axis, it must also by symmetric with respect to the origin.

93. 2 2 0x y Replace x with −x to obtain

2 2 2 20 0.x y x y The result is

the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with −y to obtain

22 2 20 0.x y x y The result is

the same as the original equation, so the graph is symmetric with respect to the x-axis. Because the graph is symmetric with respect to the x-axis and with respect to the y-axis, it must also by symmetric with respect to the origin.

94. 22 2 4x y

Replace x with −x to obtain

2 2 222 4 2 4.x y x y

The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with −y to obtain

22 2 4.x y The result is not the same

as the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with –x and y with –y to obtain

2 2 222 4 2 4,x y x y

which is not equivalent to the original equation. Therefore, the graph is not symmetric with respect to the origin.

95. To obtain the graph of ( )g x x , reflect the

graph of ( )f x x across the x-axis.

96. To obtain the graph of ( ) 2h x x , translate

the graph of ( )f x x down 2 units.



97. To obtain the graph of ( ) 2 4k x x ,

translate the graph of ( )f x x to the right 4

units and stretch vertically by a factor of 2.

98. If the graph of ( ) 3 4f x x is reflected about the x-axis, we obtain a graph whose equation is (3 4) 3 4.y x x

99. If the graph of ( ) 3 4f x x is reflected about the y-axis, we obtain a graph whose equation is ( ) 3( ) 4 3 4.y f x x x

100. If the graph of ( ) 3 4f x x is reflected about the origin, every point (x, y) will be replaced by the point (–x, –y). The equation for the graph will change from 3 4y x to

3( ) 4y x 3 4y x

3 4.y x

101. (a) To graph ( ) 3,y f x translate the graph of y = f(x), 3 units up.

(b) To graph ( 2),y f x translate the graph of y = f(x), 2 units to the right.

(c) To graph ( 3) 2,y f x translate the graph of y = f(x), 3 units to the left and 2 units down.

(d) To graph ( )y f x , keep the graph of

y = f(x) as it is where 0y and reflect the graph about the x-axis where y < 0.

102. No. For example suppose 2f x x and

2 .g x x Then

( )( ) ( ( )) (2 ) 2 2f g x f g x f x x



2 2 0 1x x . Therefore, the domain of

f g is 1, . The domain of g, ( , ), is

not a subset of the domain of , 1, .f g

For Exercises 103–110, 2( ) 3 4f x x= − and 2( ) 3 4.g x x x= − −

103. 2 2

4 3 2 2

4 3 2

( )( ) ( ) ( )

(3 4)( 3 4)

3 9 12 4 12 16

3 9 16 12 16

fg x f x g xx x x

x x x x xx x x x

104. 2 2

( )(4) (4) (4)

(3 4 4) (4 3 4 4)(3 16 4) (16 3 4 4)(48 4) (16 12 4)44 0 44

f g f g

105. 2 2

( )( 4) ( 4) ( 4)

[3( 4) 4] [( 4) 3( 4) 4][3(16) 4] [16 3( 4) 4][48 4] [16 12 4]44 24 68

f g f g

106. 2 2

2 2

2 2

2

( )(2 ) (2 ) (2 )

[3(2 ) 4] [(2 ) 3(2 ) 4]

[3(4) 4] [4 3(2 ) 4]

(12 4) (4 6 4)

16 6 8

f g k f k g kk k kk k k

k k kk k



107. 2

2

(3) 3 3 4 3 9 4(3)

(3) 9 3 3 43 3 3 427 4 23 23

9 9 4 4 4

f fg g

108.

2

2

3 1 4 3 1 4( 1)

1 3 1 41 3 1 43 4 1

undefined1 3 4 0

fg

109. The domain of (fg)(x) is the intersection of the domain of f(x) and the domain of g(x). Both have domain , , so the domain of

(fg)(x) is , .

110. 2 2

2

3 4 3 4( )

( 1)( 4)3 4

f x xxg x xx x

Because both f x and g x have domain

, , we are concerned about values of x

that make 0.g x Thus, the expression is

undefined if (x + 1)(x – 4) = 0, that is, if x = –1 or x = 4. Thus, the domain is the set of all real numbers except x = –1 and x = 4, or (– , –1) (–1, 4) (4, ).

111. 2 9f x x

( ) 2( ) 9 2 2 9f x h x h x h

( ) ( ) (2 2 9) (2 9)2 2 9 2 9 2


Thus, ( ) ( ) 2

2.f x h f x h

h h

112. 2( ) 5 3f x x x 2

2 2( ) ( ) 5( ) 3

2 5 5 3


2 2 2

2 2 2

2

( ) ( )

( 2 5 5 3) ( 5 3)

2 5 5 3 5 3

2 5

f x h f xx xh h x h x x


2( ) ( ) 2 5

(2 5)2 5

f x h f x xh h hh h

h x h x hh

For Exercises 113–118, 2( ) 2 and ( ) .f x x g x x

113. 2

( )( ) [ ( )] 2

2 2

g f x g f x g x

x x

114. 2 2( )( ) [ ( )] ( ) 2f g x f g x f x x

115. 2, so 3 3 2 1 1.f x x f

Therefore,

3 3 1g f g f g 21 1.

116. 22 , so 6 6 36.g x x g

Therefore, 6 6 36f g f g f

36 2 34 .

117. 1 1 1 2 3g f g f g g

Because 3 is not a real number, 1g f

is not defined.

118. To find the domain of ,f g we must consider the domain of g as well as the composed function, .f g Because

2 2f g x f g x x we need to

determine when 2 2 0.x Step 1: Find the values of x that satisfy

2 2 0.x 2 2 2x x

Step 2: The two numbers divide a number line into three regions.

Step 3 Choose a test value to see if it satisfies

the inequality, 2 2 0.x

Interval Test

Value Is 2 2 0x true or false?

, 2 2 22 2 0 ?2 0 True

2, 2 0 20 2 0 ?

2 0 False

2, 2 22 2 0 ?

2 0 True

The domain of f g is

, 2 2, .



119. 1 1 1 7 1 8f g f g

120. ( )(3) (3) (3) 9 9 0f g f g

121. ( )( 1) ( 1) ( 1) 3 2 6fg f g

122. (0) 5

(0) undefined(0) 0

f fg g

123. ( )( 2) [ ( 2)] (1) 2g f g f g

124. ( )(3) [ (3)] ( 2) 1f g f g f

125. ( )(2) [ (2)] (2) 1f g f g f

126. ( )(3) [ (3)] (4) 8g f g f g

127. Let x = number of yards. f(x) = 36x, where f(x) is the number of inches. g(x) = 1760x, where g(x) is the number of yards. Then

( )( ) ( ) 1760(36 )g f x g f x x 63,360 .x

There are 63,360x inches in x miles

128. Use the definition for the perimeter of a rectangle. P = length + width + length + width

( ) 2 2 6P x x x x x x This is a linear function.

129. If 343

( )V r r and if the radius is increased

by 3 inches, then the amount of volume gained is given by

3 34 43 3

( ) ( 3) ( ) ( 3) .gV r V r V r r r

130. (a) 2V r h If d is the diameter of its top, then h = d

and 2

.dr So,

32 2

2 4 4( ) ( ) ( ) .d d dV d d d

(b) 22 2S r rh

2

2 2 2

2 22 2 2

2 32 2 2

( ) 2 2 ( )d d d

d d d

S d d d

Chapter 2 Test

1. (a) The domain of 3f x x occurs

when 0.x In interval notation, this

correlates to the interval in D, 0, .

(b) The range of 3f x x is all real


(c) The domain of 2 3f x x is all real


(d) The range of 2 3f x x is all real

numbers greater than or equal to 3. In interval notation, this correlates to the interval in B, 3, .

(e) The domain of 3 3f x x is all real


(f) The range of 3 3f x x is all real


(g) The domain of 3f x x is all real


(h) The range of 3f x x is all real


(i) The domain of 2x y is 0x because when you square any value of y, the outcome will be nonnegative. In interval notation, this correlates to the interval in D, 0, .

(j) The range of 2x y is all real numbers. In interval notation, this correlates to the interval in C, , .

2. Consider the points 2,1 and 3, 4 .

4 1 3

3 ( 2) 5m

3. We label the points A 2,1 and B 3, 4 .

2 2

2 2

( , ) [3 ( 2)] (4 1)

5 3 25 9 34

d A B

Chapter 2 Test 261


4. The midpoint has coordinates

2 3 1 4 1 5, , .

2 2 2 2

5. Use the point-slope form with 3

1 1 5( , ) ( 2,1) and .x y m

1 13535

( )

1 [ ( 2)]

1 ( 2) 5 1 3( 2)5 5 3 6 5 3 11

3 5 11 3 5 11

y y m x xy xy x y xy x y x

x y x y

6. Solve 3x – 5y = –11 for y.

3 115 5

3 5 115 3 11

x yy xy x

Therefore, the linear function is 3 115 5

( ) .f x x

7. (a) The center is at (0, 0) and the radius is 2, so the equation of the circle is

2 2 4x y .

(b) The center is at (1, 4) and the radius is 1, so the equation of the circle is

2 2( 1) ( 4) 1x y

8. 2 2 4 10 13 0x y x y Complete the square on x and y to write the equation in standard form:

2 2

2 2

2 2

2 2

4 10 13 0

4 10 13

4 4 10 25 13 4 25

2 5 16

x y x yx x y y

x x y y

x y

The circle has center (−2, 5) and radius 4.

9. (a) This is not the graph of a function because some vertical lines intersect it in more than one point. The domain of the relation is [0, 4]. The range is [–4, 4].

(b) This is the graph of a function because no vertical line intersects the graph in more than one point. The domain of the function is (– , –1) (–1, ). The

range is (– , 0) (0, ). As x is getting

larger on the intervals , 1 and

1, , the value of y is decreasing, so

the function is decreasing on these intervals. (The function is never increasing or constant.)

10. Point A has coordinates (5, –3). (a) The equation of a vertical line through A

is x = 5.

(b) The equation of a horizontal line through A is y = –3.

11. The slope of the graph of 3 2y x is –3.

(a) A line parallel to the graph of 3 2y x has a slope of –3.

Use the point-slope form with

1 1( , ) (2,3) and 3.x y m

1 1( )3 3( 2)3 3 6 3 9

y y m x xy xy x y x

(b) A line perpendicular to the graph of

3 2y x has a slope of 13

because

13

3 1.

13

713 3

3 ( 2)

3 3 2 3 9 2

3 7

y xy x y x

y x y x

12. (a) 2, (b) 0, 2

(c) , 0 (d) ,

(e) , (f) 1,

13. To graph 2 1f x x , we translate the

graph of ,y x 2 units to the right and 1 unit

down.



14. ( ) 1f x x

To get y = 0, we need 0 1 1x 1 0.x To get y = 1, we need

1 1 2x 0 1.x Follow this pattern to graph the step function.

15. 12

3 if 2( )

2 if 2x

f x x x

For values of x with x < –2, we graph the horizontal line y = 3. For values of x with

2,x we graph the line with a slope of 12

and a y-intercept of (0, 2). Two points on this line are (–2, 3) and (0, 2).

16. (a) Shift f(x), 2 units vertically upward.

(b) Shift f(x), 2 units horizontally to the left.

(c) Reflect f(x), across the x-axis.

(d) Reflect f(x), across the y-axis.

(e) Stretch f(x), vertically by a factor of 2.

17. Starting with ,y x we shift it to the left 2 units and stretch it vertically by a factor of 2. The graph is then reflected over the x-axis and then shifted down 3 units.

18. 2 23 2 3x y

(a) Replace y with –y to obtain 2 2 2 23 2( ) 3 3 2 3.x y x y

The result is the same as the original equation, so the graph is symmetric with respect to the x-axis.

(b) Replace x with –x to obtain 2 2 2 23( ) 2 3 3 2 3.x y x y

The result is the same as the original equation, so the graph is symmetric with respect to the y-axis.

(c) The graph is symmetric with respect to the x-axis and with respect to the y-axis, so it must also be symmetric with respect to the origin.

Chapter 2 Test 263


19. 2( ) 2 3 2, ( ) 2 1f x x x g x x

(a)

2

2

2

( )( ) ( ) ( )

2 3 2 2 1

2 3 2 2 1

2 1

f g x f x g xx x x

x x xx x

(b) 2( ) 2 3 2

( )( ) 2 1

f f x x xxg g x x

(c) We must determine which values solve the equation 2 1 0.x

12

2 1 0 2 1x x x

Thus, 12

is excluded from the domain,

and the domain is 1 12 2

, , .

(d) 2( ) 2 3 2f x x x

2

2 2

2 2

2 3 2

2 2 3 3 2

2 4 2 3 3 2


x xh h x h

2 2

2

2 2

2

2

( ) ( )

(2 4 2 3 3 2)

(2 3 2)

2 4 2 3

3 2 2 3 2

4 2 3

f x h f xx xh h x h

x xx xh h x

h x xxh h h

2( ) ( ) 4 2 3

(4 2 3)

4 2 3

f x h f x xh h hh h

h x hh

x h

(e) 2

( )(1) (1) (1)

(2 1 3 1 2) ( 2 1 1)(2 1 3 1 2) ( 2 1 1)(2 3 2) ( 2 1)1 ( 1) 0

f g f g

(f) 2

( )(2) (2) (2)

(2 2 3 2 2) ( 2 2 1)(2 4 3 2 2) ( 2 2 1)(8 6 2) ( 4 1)4( 3) 12

fg f g

(g) 2 1 0 2 0 1g x x g

0 1 1. Therefore,

2

0 0

1 2 1 3 1 22 1 3 1 22 3 2 1

f g f gf

For exercises 20 and 21, 1f x x and

2 7.g x x

20. 2 7

(2 7) 1 2 6

f g f g x f xx x


the domain of f is [0, ) . We need to find the values of x which fit the domain of f: 2 6 0 3x x . So, the domain of f g

is [3, ) .

21. 1

2 1 7

g f g f x g x

x


the domain of f is [0, ) . We need to find the values of x which fit the domain of f:

1 0 1x x . So, the domain of g f

is [ 1, ) .

22. (a) C(x) = 3300 + 4.50x

(b) R(x) = 10.50x

(c) ( ) ( ) ( )10.50 (3300 4.50 )6.00 3300

P x R x C xx x

x

(d) ( ) 06.00 3300 0

6.00 3300550

P xx

xx

She must produce and sell 551 items before she earns a profit.

20 Copyright © 2017 Pearson Education, Inc

Chapter 2 Graphs and Functions Section 2.1 Rectangular Coordinates

and Graphs Classroom Example 1 (page 184)

(a) (transportation, $12,153)

(b) (health care, $4917)


22( , ) 2 3 8 5

25 169 194

d P Q


22

22

2 2

( , ) 5 0 1 2

25 9 34

( , ) 4 0 3 2

16 25 41

( , ) 4 5 3 1

81 4 85

d R S

d R T

d S T

The longest side has length 85

2 2 2?34 41 85

34 41 85

The triangle formed by the three points is not a right triangle.


2 2

2 2

22

The distance between ( 2,5) and (0,3) is

2 0 5 3 4 4 8 2 2

The distance between (0,3) and (8, 5) is

8 0 5 3 64 64

128 8 2

The distance between ( 2,5) and (8, 5) is

2 8 5 5 100 100

200 10 2

P Q

Q R

P R

Because 2 2 8 2 10 2 , the points are collinear.


(a) The coordinates of M are

7 ( 2) 5 13 9

, , 42 2 2

(b) Let (x, y) be the coordinates of Q. Use the midpoint formula to find the coordinates:

8 20, 4, 4

2 2

84 8 8 0

220

4 20 8 122

x y

xx x

yy y

The coordinates of Q are (0, 12).


The year 2011 lies halfway between 2009 and 2013, so we must find the coordinates of the midpoint of the segment that has endpoints (2009, 124.0) and (2013, 137.4)

2009 2013 124.0 137.4,

2 2

2011, 130.7

M

The estimate of $130.7 billion is $0.1 billion more than the actual amount.


Choose any real number for x, substitute the value in the equation and then solve for y. Note that additional answers are possible.

(a) x y = −2x + 5

−1 2( 1) 5 7y

0 2(0) 5 5y

3 2(3) 5 1y

Three ordered pairs that are solutions are (−1, 7), (0, 5), and (3, −1). Other answers are possible.



(b) y 3 1x y

−9 3 39 1 8 2x

−2 3 32 1 1 1x

−1 3 31 1 0 0x

0 33 0 1 1 1x

7 3 37 1 8 2x

Ordered pairs that are solutions are (−2, −9), (−1, −2), (0, −1), (1, 0) and (2, 7). Other answers are possible.

(c) x 2 1y x

−2 2( 2) 1 3y

−1 2( 1) 1 0y

0 20 1 1y

1 21 1 0y

2 22 1 3y

Ordered pairs that are solutions are (−2, −3), (−1, 0), (0, 1), (1, 0) and (2, −3). Other answers are possible.


(a) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:

52

0 2 52(0) 5 5x x

y y

Find a third point on the graph by letting x = −1 and solving for y: 2 1 5 7y .

The three points are 52

, 0 , (0, 5), and (−1, 7).

Note that (3, −1) is also on the graph.

(b) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:

33

3

0 1 1 1

0 1 0 1 1

x

y y y


for y: 332 1 2 1 7y y y .

Find a fourth point by letting x = −2 and solving for y:

332 1 2 1 9y y y

The points to be plotted are (0, −1), (1, 0), (2, 7), and (−2, −9). Note that (−1, −2) is also on the graph.

(c) Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept:

2 2 2

2

0 1 1 1 1

0 1 1

x x x x

y


for y: 22 1 3y . Find a fourth point

by letting x = −2 and solving for y:

22 1 3y

The points to be plotted are (−1, 0), (1, 0), (0, 1), (2, −3), and (−2, −3)

Section 2.2 Circles Classroom Example 1 (page 195)

(a) (h, k) = (1, −2) and r = 3

2 2 2

22 2

2 2

1 2 3

1 2 9

x h y k r

x y

x y

(b) (h, k) = (0, 0) and r = 2

2 2 2

2 2 2

2 2

0 0 2

4

x h y k r

x y

x y




(a) 2 21 2 9x y

This is a circle with center (1, −2) and radius 3.

(b) 2 2 4x y

This is a circle with center (0, 0) and radius 2.



2 2

2 2

2 2

4 8 44 0

4 4 8 16 44 4 16

2 4 64

x x y y

x x y y

x y

Because c = 64 and 64 > 0, the graph is a circle. The center is (−2, 4) and the radius is 8.


2 22 2 2 6 45x y x y

Group the terms, factor out 2, and then complete the square:

2 21 92 2 3

4 4

1 945 2 2

4 4

x x y y

Factor and then divide both sides by 2:

2 2

2 2

1 32 2 50

2 2

1 325

2 2

x y

x y

Because c = 25 and 25 > 0, the graph is a circle. The

center is 1 3

,2 2

and the radius is 5.



2 2

2 2

2 2

6 2 12 0

6 9 2 1 12 9 1

3 1 2

x x y y

x x y y

x y

Because c = −2 and −2 < 0, the graph is nonexistent.


Determine the equation for each circle and then substitute x = −3 and y = 4. Station A:

2 2 2

2 2 2

2 2

1 4 4

3 1 4 4 4

4 416 16

x y

Station B:

2 2 2

2 2

2 2

2 2

6 0 5

6 25

3 6 4 25

3 4 259 16 25

25 25

x y

x y

Station C:

22 2

2 2

2 2

2 2

5 2 10

5 2 100

3 5 4 2 100

8 6 10064 36 100

100 100

x y

x y

Because (−3, 4) satisfies all three equations, we can conclude that the epicenter is (−3, 4).



Section 2.3 Functions Classroom Example 1 (page 204)

4,0 , 3,1 , 3,1M

M is a function because each distinct x value has exactly one y value.

2,3 , 3, 2 , 4,5 , 5, 4N

N is a function because each distinct x value has exactly one y value.

4,3 , 0,6 , 2,8 , 4, 3P

P is not a function because there are two y-values for x = −4.


(a) Domain: {−4, −1, 1, 3} Range: {−2, 0, 2, 5} The relation is a function.

(b) Domain: {1, 2, 3} Range: {4, 5, 6, 7} The relation is not a function because 2 maps to 5 and 6.

(c) Domain: {−3, 0, 3, 5} Range: {5} The relation is a function.


(a) Domain: {−2, 4}; range: {0, 3}

(b) Domain: ( , ) ; range: ( , )

(c) Domain: [−5, 5]; range: [−3, 3]

(d) Domain: ( , ) ; range: ( , 4]


(a)

Not a function

(b)

Function

(c)

Not a function

(d)

Function


(a) 2 5y x represents a function because y is

always found by multiplying x by 2 and subtracting 5. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers or ( , ) .

Because y is twice x, less 5, y also may be any real number, and so the range is also all real numbers, ( , ) .



(b) For any choice of x in the domain of 2 3y x , there is exactly one corresponding

value for y, so the equation defines a function. The function is defined for all values of x, so the domain is ( , ) . The square of any

number is always positive, so the range is [3, ) .

(c) For any choice of x in the domain of x y ,

there are two possible values for y. Thus, the equation does not define a function. The domain is [0, ) while the range is ( , ) .

(d) By definition, y is a function of x if every value of x leads to exactly one value of y. Substituting a particular value of x, say 1, into y x corresponds to many values of y. The

ordered pairs (0, 2) (1, 1) (1, 0) (3, −1) and so on, all satisfy the inequality. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers, ( , ) .

(e) For 3

2y

x

, we see that y can be found by

dividing x + 2 into 3. This process produces one value of y for each value of x in the domain. The domain includes all real numbers except those that make the denominator equal to zero, namely x = −2. Therefore, the domain is ( , 2) ( 2, ) . Values of y can be

negative or positive, but never zero. Therefore the range is ( , 0) (0, ) .


(a) 2( 3) ( 3) 6( 3) 4 13f

(b) 2( ) 6 4f r r r

(c) ( 2) 3( 2) 1 3 7g r r r


(a) 2( 1) 2( 1) 9 7f

(b) ( 1) 6f

(c) ( 1) 5f

(d) ( 1) 0f


(a) 2

2

2

( ) 2 3

( 5) ( 5) 2( 5) 3 12

( ) 2 3

f x x x

f

f t t t

(b) 2

2 3 6 23

2( ) 2

32 16

( 5) ( 5) 23 3

2( ) 2

3

x y y x

f x x

f

f t t




The function is increasing on ( , 1), decreasing

on (–1, 1) and constant on (1, ).


The example refers to the following figure.

(a) The water level is changing most rapidly from 0 to 25 hours.

(b) The water level starts to decrease after 50 hours.

(c) After 75 hours, there are 2000 gallons of water in the pool.

Section 2.4 Linear Functions Classroom Example 1 (page 220)

32

( ) 6f x x ; Use the intercepts to graph the

function. 32


3 32 2

0 6 6 4 :x x x x-intercept



( ) 2f x is a constant function. Its graph is a

horizontal line with a y-intercept of 2.

Domain: ( , ), range: {2}


x = 5 is a vertical line intersecting the x-axis at (5, 0).

Domain:{5}, range: ( , )


3 4 0;x y Use the intercepts.

3 0 4 0 4 0 0 : -intercepty y y y

3 4 0 0 3 0 0 : -interceptx x x x


3 4 4 0 12 4 04 12 3

y yy y

Graph the line through (0, 0) and (4, −3).





(a) 4 ( 6) 10 5

2 2 4 2m

(b) 8 8 0

05 ( 3) 8

m

(c) 10 10 20

4 ( 4) 0m

the slope is

undefined.


2x − 5y = 10 Solve the equation for y. 2 5 10

5 2 102

25

x yy x

y x

The slope is 2

,5

the coefficient of x.


First locate the point (−2, −3). Because the slope is 43

, a change of 3 units horizontally (3 units to the

right) produces a change of 4 units vertically (4 units up). This gives a second point, (1, 1), which can be used to complete the graph.


The average rate of change per year is 8658 9547 889

296.33 million2013 2010 3

The graph confirms that the line through the ordered pairs fall from left to right, and therefore has negative slope. Thus, the amount spent by the federal government on R&D for general science decreased by an average of $296.33 million (or $296,330,000) each year from 2010 to 2013.


(a) ( ) 120 2400C x x

(b) ( ) 150R x x

(c) ( ) ( ) ( )150 (120 2400)30 2400

P x R x C xx x

x

(d) ( ) 0 30 2400 0 80P x x x

At least 81 items must be sold to make a profit.



( 5) 2( 3)5 2 6

2 1

y xy x

y x


First find the slope: 3 ( 1) 4

4 5 9m

Now use either point for 1 1( , )x y : 49

3 [ ( 4)]9( 3) 4( 4)9 27 4 16

9 4 114 9 11

y xy xy x

y xx y


Write the equation in slope-intercept form: 34

3 4 12 4 3 12 3x y y x y x

The slope is 34

, and the y-intercept is (0, −3).




First find the slope: 4 2 1

2 2 2m

Now, substitute 12

for m and the coordinates of one

of the points (say, (2, 2)) for x and y into the slope-intercept form y = mx + b, then solve for b:

12

2 2 3b b . The equation is

12

3y x .


The example refers to the following figure:

(a) The line rises 5 units each time the x-value

increases by 2 units. So the slope is 52

. The

y-intercept is (0, 5), and the x-intercept is (−2, 0).


( ) 5f x x .


(a) Rewrite the equation 3 2 5x y in slope-

intercept form to find the slope: 3 52 2

3 2 5x y y x The slope is 32

.

The line parallel to the equation also has slope 32

. An equation of the line through (2, −4) that

is parallel to 3 2 5x y is 3 32 2

( 4) ( 2) 4 3y x y x

32

7y x or 3x − 2y = 14.

(b) The line perpendicular to the equation has

slope 23

. An equation of the line through

(2, −4) that is perpendicular to 3 2 5x y is 2 2 43 3 3

( 4) ( 2) 4y x y x 82

3 3y x or 2x +3y = −8.


(a) First find the slope: 7703 6695

5043 1

m

Now use either point for 1 1( , )x y :

6695 504( 1)6695 504 504

504 6191

y xy x

y x

(b) The year 2015 is represented by x = 6. 504(6) 6191 9215y

According to the model, average tuition and fees for 4-year colleges in 2015 will be about $9215.


Write the equation as an equivalent equation with 0 on one side.

3 2(5 ) 2 38x x x

3 2(5 ) 2 38 0x x x

Now graph the equation to find the x-intercept.

The solution set is {−4}.

Section 2.6 Graphs of Basic Functions Classroom Example 1 (page 249)

(a) The function is continuous over ( , 0) (0, )

(b) The function is continuous over its entire domain ( , ).




(a) Graph each interval of the domain separately. If x < 1, the graph of f(x) = 2x + 4 has an endpoint at (1, 6), which is not included as part of the graph. To find another point on this part of the graph, choose x = 0, so y = 4. Draw the ray starting at (1, 6) and extending through (0, 4). Graph the function for x ≥ 1, f(x) = 4 − x similarly. This part of the graph has an endpoint at (1, 3), which is included as part of the graph. Find another point, say (4, 0), and draw the ray starting at (1, 3) which extends through (4, 0).

(b) Graph each interval of the domain separately. If x ≤ 0, the graph of f(x) = −x − 2 has an endpoint at (0, −2), which is included as part of the graph. To find another point on this part of the graph, choose x = −2, so y = 0. Draw the ray starting at (0, −2) and extending through (−2, 0). Graph the function for x > 0,

2( ) 2f x x similarly. This part of the graph

has an endpoint at (0, −2), which is not included as part of the graph. Find another point, say (2, 2), and draw the curve starting at (0, −2) which extends through (2, 2). Note that the two endpoints coincide, so (0, −2) is included as part of the graph.


Create a table of sample ordered pairs:

x −6 −3 32

0 32

3 6

13

2y x −4 −3 −3 −2 −2 −1 0


For x in the interval (0, 2], y = 20. For x in (2, 3], y = 20 + 2 = 22. For x in (3, 4], y = 22 + 2 = 24. For x in (4, 5], y = 24 + 2 = 26. For x in (5, 6], y = 26 + 2 = 28.

Section 2.7 Graphing Techniques Classroom Example 1 (page 260)

Use this table of values for parts (a)−(c)

x 2( ) 2g x x 212

( )h x x 212

( )k x x

−2 8 2 1

−1 2 12

14

0 0 0 0

1 2 12

14

2 8 2 1

(a) 2( ) 2g x x



(b) 212

( )h x x

(c) 2

1( )

2k x x


Use this table of values for parts (a) and (b)

x ( )g x x ( )h x x

−2 −2 2

−1 −1 1

0 0 0

1 −1 1

2 −2 2

(a) ( )g x x

(b) ( )h x x


(a) x y

Replace x with –x to obtain x y . The result is not the same as the

original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain x y x y The

result is the same as the original equation, so the graph is symmetric with respect to the x-axis. The graph is symmetric with respect to the x-axis only.

(b) 3y x

Replace x with –x to obtain 3y x

3y x . The result is the same as the

original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain 3y x . The result is not the same

as the original equation, so the graph is not symmetric with respect to the x-axis. Therefore, the graph is symmetric with respect to the y-axis only.

(c) 2 6x y



original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain 2 ( ) 6x y


original equation, so the graph is not symmetric with respect to the x-axis. Therefore, the graph is not symmetric with respect to either axis.

(d) 2 2 25x y

Replace x with –x to obtain 2 2( ) 25x y 2 2 25x y . The result is

the same as the original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain

2 2( ) 25x y 2 2 25x y . The result is

the same as the original equation, so the graph is symmetric with respect to the x-axis. Therefore, the graph is symmetric with respect to both axes. Note that the graph is a circle of radius 5, centered at the origin.




(a) 32y x

Replace x with –x and y with –y to obtain 3 3 3( ) 2( ) 2 2y x y x y x . The

result is the same as the original equation, so the graph is symmetric with respect to the origin.

(b) 22y x

Replace x with –x and y with –y to obtain 2 2 2( ) 2( ) 2 2y x y x y x .

The result is not the same as the original equation, so the graph is not symmetric with respect to the origin.


(a) 5 3( ) 2 3g x x x x

Replace x with –x to obtain 5 3

5 3

5 3

( ) ( ) 2( ) 3( )

2 3

( 2 3 ) ( )

g x x x x

x x x

x x x g x

g(x) is an odd function.

(b) 2( ) 2 3h x x

Replace x with –x to obtain 2 2( ) 2( ) 3 2 3 ( )h x x x h x h(x) is

an even function.

(c) 2( ) 6 9k x x x

Replace x with –x to obtain 2

2( ) ( ) 6( ) 9

6 9 ( ) and ( )

k x x x

x x k x k x

k(x) is neither even nor odd.


Compare a table of values for 2( )g x x with 2( ) 2f x x . The graph of f(x) is the same as the

graph of g(x) translated 2 units up.

x 2( )g x x 2( ) 2f x x

−2 4 6

−1 1 3

0 0 2

1 1 3

2 4 6


Compare a table of values for 2( )g x x with 2( ) ( 4)f x x . The graph of f(x) is the same as the

graph of g(x) translated 4 units left.

x 2( )g x x 2( ) ( 4)f x x

−7 49 9

−6 36 4

−5 25 1

−4 16 0

−3 9 1

−2 4 4

−1 1 9


(a) 2( ) ( 1) 4f x x

This is the graph of 2( )g x x , translated one

unit to the right, reflected across the x-axis, and then translated four units up.

x 2( )g x x 2( ) ( 1) 4f x x

−2 4 −5

−1 1 0

0 0 3

1 1 4

2 4 3

3 9 0

4 16 −5




(continued)

(b) ( ) 2 3f x x

This is the graph of ( )g x x , translated three

units to the left, reflected across the x-axis, and then stretched vertically by a factor of two.

x ( )g x x ( ) 2 3f x x

−6 6 −6

−5 5 −4

−4 4 −2

−3 3 0

−2 2 −2

−1 1 −4

0 0 −6

(c) 12

( ) 2 3h x x

This is the graph of ( )g x x , translated two

units to the left, shrunk vertically by a factor of 2, and then translated 3 units down.

x ( )g x x 12

( ) 2 3h x x

−2 undefined −3

−1 undefined −2.5

0 0 12

2 3 2.3

2 2 1.4 −2

6 6 2.4 12

8 3 1.6

7 7 2.6 −1.5


The graphs in the exercises are based on the following graph.

(a) ( ) ( ) 2g x f x

This is the graph of f(x) translated two units down.

(b) ( ) ( 2)h x f x

This is the graph of f(x) translated two units right.



(c) ( ) ( 1) 2k x f x

This is the graph of f(x) translated one unit left, and then translated two units up.

(d) ( ) ( )F x f x

This is the graph of f(x) reflected across the y-axis.



For parts (a)−(d), ( ) 3 4f x x and 2( ) 2 1g x x

(a) (0) 3(0) 4 4f and 2(0) 2(0) 1 1g , so

( )(0) 4 1 5f g

(b) (4) 3(4) 4 8f and 2(4) 2(4) 1 31g ,

so ( )(4) 8 31 23f g

(c) ( 2) 3( 2) 4 10f and 2( 2) 2( 2) 1 7g , so

( )( 2) ( 10)(7) 70fg

(d) (3) 3(3) 4 5f and 2(3) 2(3) 1 17g ,

so 5

(3)17

f

g


For parts (a)−(e), 2( ) 3f x x x and ( ) 4 5g x x

(a) 2 2( )( ) ( 3 ) (4 5) 5f g x x x x x x

(b) 2 2( )( ) ( 3 ) (4 5) 7 5f g x x x x x x

(c) 2

3 2 2

3 2

( )( ) ( 3 )(4 5)

4 5 12 15

4 7 15

fg x x x x

x x x x

x x x

(d) 2 3

( )4 5

f x xx

g x

(e) The domains of f and g are both ( , ) . So,

the domains of f + g, f − g, and fg are the intersection of the domains of f and g,

( , ) . The domain of fg

includes those

real numbers in the intersection of the domains of f and g for which ( ) 4 5 0g x x

54

x . So the domain of fg

is

5 54 4

, , .


(a)

From the figure, we have f(1) = 3 and

g(1) = 1, so (f + g)(1) = 3 + 1 = 4. f(0) = 4 and g(0) = 0, so (f − g)(0) = 4 − 0 = 4. f(−1) = 3 and g(−1) = 1, so (f g)(−1) = (3)(1) = 3

f(−2) = 0 and g(−2) = 2, so 02

( 2) 0fg

.

(b) x f(x) g(x)

−2 −5 0

−1 −3 2

0 −1 4

1 1 6

From the table, we have f(1) = 1 and g(1) = 6, so (f + g)(1) = 1 + 6 = 7. f(0) = −1 and g(0) = 4, so (f − g)(1) = −1 − 4 = −5. f(−1) = −3 and g(−1) = 2, so (f g)(−1) = (−3)(2) = −6 f(−2) = −5 and g(−2) = 0, so

50

( 2)fg

fg

is undefined.



(c) ( ) 3 4, ( )f x x g x x

From the formulas, we have (1) 3(1) 4 7f and (1) 1 1g , so

(f + g)(1) = 7 + (−1) = 6. (0) 3(0) 4 4f and (0) 0 0g , so

(f − g)(1) = 4 − 0 = 4. ( 1) 3( 1) 4 1f and ( 1) 1 1g ,

so (fg)(−1) =(1)(−1) = −1. ( 2) 3( 2) 4 2f and

( 2) 2 2g , so 2

( 2) 1.2

f

g


Step 1: Find ( )f x h : 2

2 2

2 2

( ) 3( ) 2( ) 4

3( 2 ) 2 2 4

3 6 3 2 2 4

f x h x h x h

x xh h x h

x xh h x h

Step 2: Find ( ) ( )f x h f x :

2 2 2

2

( ) ( )

(3 6 3 2 2 4) (3 2 4)

6 3 2

f x h f x

x xh h x h x x

xh h h

Step 3: Find the difference quotient: 2( ) ( ) 6 3 2

6 3 2f x h f x xh h h

x hh h


For parts (a) and (b), ( ) 4f x x and 2

( )g xx

(a) First find g(2): 2

(2) 12

g . Now find

( )(2) ( (2)) (1) 1 4 5f g f g f

(b) First find f(5): (5) 5 4 9 3f . Now

find 23

( )(5) ( (5)) (3)g f g f g

The screens show how a graphing calculator

evaluates the expressions in this classroom example.


For parts (a) and (b), ( ) 1f x x and

( ) 2 5g x x

(a) ( )( ) ( ( )) (2 5) 1 2 4f g x f g x x x

The domain and range of g are both ( , ) .

However, the domain of f is [1, ) . Therefore,

( )g x must be greater than or equal to 1:

2 5 1 2x x . So, the domain of f g

is [ 2, ) .

(b) ( )( ) ( ( )) 2 1 5g f x g f x x

The domain of f is [1, ) , while the range of f

is [0, ) . The domain of g is ( , ) .

Therefore, the domain of ( )g f is restricted

to that portion of the domain of g that intersects with the domain of f, that is [1, ) .


For parts (a) and (b), 5

( )4

f xx

and 2

( )g xx

(a) 5 5

( )( ) ( ( ))2 4 2 4

xf g x f g x

x x

The domain and range of g are both all real numbers except 0. The domain of f is all real numbers except −4. Therefore, the expression for ( )g x cannot equal −4. So,

2 14

2x

x . So, the domain of f g

is the set of all real numbers except for 12

,

and 0. This is written

1 12 2

, , 0 0,



(b) 2 2 8

( )( ) ( ( ))5 ( 4) 5

xg f x g f x

x

The domain of f is all real numbers except −4, while the range of f is all real numbers except 0. The domain and range of g are both all real numbers except 0, which is not in the range of f. So, the domain of g f is the set of all real

numbers except for −4. This is written , 4 4,


( ) 2 5f x x and 2( ) 3g x x x 2

2

2

2 2

2

( )( ) (2 5) 3(2 5) (2 5)

3(4 20 25) 2 5

12 58 70

( )( ) (3 ) 2(3 ) 5

6 2 5

g f x g x x x

x x x

x x

f g x f x x x x

x x

In general, 2 212 58 70 6 2 5,x x x x so

( )( ) ( )( )g f x f g x .


2( )( ) 4(3 2) 5(3 2) 8f g x x x

Answers may vary. Sample answer: 2( ) 4 5 8f x x x and ( ) 3 2g x x .

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